Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9

Direction (1 – 20): Evaluate the definite integrals.

Question 1.
\(\int_{-1}^{1}\) (x + 1) dx
Solution.
Let I = \(\int_{-1}^{1}\) (x + 1) dx
∫ (x + 1) dx = \(\frac{x^{2}}{2}\) + x = F(x)
By second fundamental theorem of calculus, we get
I = F(1) – F(- 1)
= \(\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)\)

= \(\frac{1}{2}\) + 1 – \(\frac{1}{2}\) + 1 = 2.

Question 2.
\(\int_{2}^{3} \frac{1}{x}\) dx
Solution.
Let I = \(\int_{2}^{3} \frac{1}{x}\) dx
∫ \(\frac{1}{x}\) dx = log |x| = F(x)
By second fundamental theorem of calculus, we get
I = F(3) – F(2)
= log |3| – log |2| [∵ log(a) – log(b) = log (\(\frac{a}{b}\))]
= log \(\frac{3}{2}\)

Question 3.
\(\int_{1}^{2}\) (4x³ + 5x² + 6x + 9) dx
Solution.
Let I = \(\int_{1}^{2}\) (4x³ + 5x² + 6x + 9) dx
∫ (4x³ + 5x² + 6x + 9) dx = \(4\left(\frac{x^{4}}{4}\right)-5\left(\frac{x^{3}}{3}\right)+6\left(\frac{x^{2}}{2}\right)\) + 9(x)
= x⁴ – \(\frac{5 x^{3}}{3}\) + 3x² + 9x = F(x)
By second fundamental theorem of calculus, we get
I = F(2) – F(1)

Question 4.
\(\int_{0}^{\frac{\pi}{4}}\) sin 2x dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{4}}\) sin 2x dx

∫ sin 2x dx = \(\left(\frac{-\cos 2 x}{2}\right)\) = F(x)
By second fundamental theorem of calculus, we get
I = F(\(\frac{\pi}{4}\)) – F(0)

= \(\frac{1}{2}\left[\cos 2\left(\frac{\pi}{4}\right)-\cos 0\right]\)

= – \(\frac{1}{2}\) [0 – 1]

= \(\frac{1}{2}\)

Question 5.
\(\int_{0}^{\frac{\pi}{2}}\) cos 2x dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}}\) cos 2x dx

∫ cos 2x = \(\left(\frac{\sin 2 x}{2}\right)\) = F(x)

By second fundamental theorem of calculus, we get
I = F(\(\frac{\pi}{4}\)) – F(0)

= \(\frac{1}{2}\) [sin 2(\(\frac{\pi}{2}\)) – sin 0]

= \(\frac{1}{2}\) [sin π – sin 0]

= \(\frac{1}{2}\) [0 – 0] = 0.

Question 6.
\(\int_{4}^{5}\) e^x dx
Solution.
Let I = \(\int_{4}^{5}\) e^x dx
∫ e^x dx = ex = F(x)
By second fundamental theorem of calculus, we get
I = F(5) – F(4)
= e⁵ – e⁴
= e⁴ (e – 1)

Question 7.
\(\int_{0}^{\frac{\pi}{4}}\) tan x dx
Sol.
Let I = \(\int_{0}^{\frac{\pi}{4}}\) tan x dx
∫ tan x dx = – log |cos x| =F(x)
By second fundamental theorem of calculus, we get
I = F(\(\frac{\pi}{4}\)) – F(0)

= – log |cos \(\frac{\pi}{4}\)| + log |cos 0|

= – log |\(\frac{1}{\sqrt{2}}\)| + log |1|

= – log \((2)^{-\frac{1}{2}}\)

= \(\frac{1}{2}\) log 2.

Question 8.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) cosec x dx
Solution.
21et I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) cosec x dx
∫ cosec x dx = log cosec x – cot x = F(x)
By second fundamental theorem of calculus, we get
I = F(\(\frac{\pi}{4}\)) – F(\(\frac{\pi}{6}\))

= log |cosec \(\frac{\pi}{4}\) – cot \(\frac{\pi}{4}\)| – log |cosec \(\frac{\pi}{6}\) – cot \(\frac{\pi}{6}\)|

= log |√2 – 1| – log |2 – √3|

= log \(\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
\(\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\)
Solution.
Let I = \(\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\)

⇒ ∫ \(\frac{d x}{\sqrt{1-x^{2}}}\) = sin^-1 x = F(x)

By second fundamental theorem of calculus, we get
I = F(1) – F(0)
= sin^-1 (1) – sin^-1 (0)

= \(\frac{\pi}{2}\) – 0 = \(\frac{\pi}{2}\)

Question 10.
\(\int_{0}^{1} \frac{d x}{1+x^{2}}\)
Solution.
Let I = \(\int_{0}^{1} \frac{d x}{1+x^{2}}\)

∫ \(\frac{d x}{1+x^{2}}\) = tan^-1 x = F(x)
By second fundamental theorem of calculus, we get
I = F(1) – F(0)
= tan^-1 (1) – tan^-1 (0) = \(\frac{\pi}{4}\)

Question 11.
\(\int_{2}^{3} \frac{d x}{x^{2}-1}\)
Solution.

Question 12.
\(\cdot \int_{0}^{\frac{\pi}{2}}\) cos² x dx
Solution.
Let I = \(\cdot \int_{0}^{\frac{\pi}{2}}\) cos² x dx

Question 13.
\(\int_{2}^{3} \frac{x d x}{x^{2}+1}\)
Solution.
Let I = \(\int_{2}^{3} \frac{x d x}{x^{2}+1}\)

\(\int \frac{x}{x^{2}+1} d x=\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x\)

= \(\frac{1}{2}\) log (1 + x²) = F(x)

By second fundamental theorem of calculus, we get

I = F(3) – F(2)
= \(\frac{1}{2}\) [log(1 +(3)²) – log(1 + (2)²)]

= \(\frac{1}{2}\) [log(10) – log(5)]

= \(\frac{1}{2}\) log (\(\frac{10}{5}\))

= \(\frac{1}{2}\) log 2

Question 14.
\(\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}\) dx
Solution.
Let I = \(\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}\) dx

Question 15.
\(\int_{0}^{1}\) x e^x2 dx
Solution.
Let I = \(\int_{0}^{1}\) x e^x2 dx
put x² = t
⇒ 2x dx = dt
As x → 0, t → 0 and as x → 1, t → 1
I = \(\frac{1}{2}\) \(\int_{0}^{1}\) e^t dt

⇒ \(\frac{1}{2}\) ∫ e^t dt = \(\frac{1}{2}\) e^t = F(t)

By second fundamental theorem of calculus, we get
I = F(1) – F(0)
= \(\frac{1}{2}\) e – \(\frac{1}{2}\) e⁰

= \(\frac{1}{2}\) (e – 1)

Question 16.
\(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3}\) dx
Solution.
Let I = \(\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3}\) dx
Dividing 5x² by x² + 4x + 3, we get

Question 17.
\(\int_{0}^{\frac{\pi}{4}}\) (2 sec² x + x³ + 2) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{4}}\) (2 sec² x + x³ + 2) dx

∫ (2 sec² x + x³ + 2) dx = 2 tan x + \(\frac{x^{4}}{4}\) + 2x

By second fundamental theorem of calculus, we get

I = F(\(\frac{\pi}{4}\)) – F(0)

= \(\left\{\left(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^{4}+2\left(\frac{\pi}{4}\right)\right)\right\}\) – (2 tan 0 + 0 + 0)

= \(2 \tan \frac{\pi}{4}+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2}=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}\)

Question 18.
\(\int_{0}^{\pi}\) (sin² \(\frac{x}{2}\) – cos² \(\frac{x}{2}\)) dx
Solution.
Let I = \(\int_{0}^{\pi}\) (sin² \(\frac{x}{2}\) – cos² \(\frac{x}{2}\)) dx

= – \(\int_{0}^{\pi}\) (cos² \(\frac{x}{2}\) – sin² \(\frac{x}{2}\)) dx

= – \(\int_{0}^{\pi}\) cos x dx
∫ cos dx = sin x = F(x)
By second fundamental theorem of calculus, we get
I = F(π) – F(0)
= sin π – sin 0 = 0.

Question 19.
\(\int_{0}^{2} \frac{6 x+3}{x^{2}+4}\) dx
Solution.
Let I = \(\int_{0}^{2} \frac{6 x+3}{x^{2}+4}\) dx

Question 20.
\(\int_{0}^{1}\) (xe^x + sin \(\frac{\pi x}{4}\)) dx
Solution.
Let I = \(\int_{0}^{1}\) (xe^x + sin \(\frac{\pi x}{4}\)) dx

Question 21.
\(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\) equals
(A) \(\frac{\pi}{3}\)

(B) \(\frac{2 \pi}{3}\)

(C) \(\frac{\pi}{6}\)

(D) \(\frac{\pi}{12}\)
Solution.
Let I = \(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\)
= tan^-1 x = F(x)
By second fundamental theorem of calculus, we get
\(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\) = F(√3) – F(1)

= tan^-1 (√3) – tan^-1 (1)

= \(\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\)

Hence, the correct answer is (D).

Question 22.
\(\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}\) equals
(A) \(\frac{\pi}{6}\)

(B) \(\frac{\pi}{12}\)

(C) \(\frac{\pi}{24}\)

(D) \(\frac{\pi}{4}\)
Solution.

Hence, the correct answer is (C).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.11 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11

Direction (1 – 19):
By using the properties of definite integrals, evaluate the integrals.

Question 1.
\(\int_{0}^{\frac{\pi}{2}}\) cos x dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}}\) cos x dx ……….(i)

Question 2.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
Solution.

Question 3.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)
Solution.

Adding equations (i) and (ii), we get

⇒ 2I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\) dx

⇒ 2I = \(\int_{0}^{\frac{\pi}{2}}\) 1 dx

⇒ 2I = \([x]_{0}^{\frac{\pi}{2}}\)

⇒ 2I = \(\frac{\pi}{2}\)

⇒ I =\(\frac{\pi}{4}\)

Question 4.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x}\) dx
Solution.

Question 5.
\(\int_{-5}^{5}\) |x + 2| dx
Solution.
Let I = \(\int_{-5}^{5}\) |x + 2| dx
It can be seen that (x + 2) ≤ 0 on [- 5, – 2]and (x + 2) ≥ 0 on [- 2, 5].

Question 6.
\(\int_{2}^{8}\) |x – 5| dx
Solution.
Let I = \(\int_{2}^{8}\) |x – 5| dx
It can be seen that (x – 5) ≤ 0 on [2, 5] and (x – 5) ≥ 0 on [5, 8].

Question 7.
\(\int_{0}^{1}\) x (1 – x)ⁿ dx
Solution.
Let I = \(\int_{0}^{1}\) x (1 – x)ⁿ dx
I = \(\int_{0}^{1}\) (1 – x) (1 – (1 – x))ⁿ dx
= \(\int_{0}^{1}\) (1 – x) (x)ⁿ dx
= \(\int_{0}^{1}\) (xⁿ – xⁿ⁺¹) dx
= \(\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}\)

(∵ \(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx

= \(\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\)

= \(\frac{(n+2)-(n+1)}{(n+1)(n+2)}\)

= \(\frac{1}{(n+1)(n+2)}\)

Question 8.
\(\int_{0}^{\frac{\pi}{4}}\) log (1 + tan x) dx
Solution.

⇒ I = \(\int_{0}^{\frac{\pi}{4}}\) log 2 dx – \(\int_{0}^{\frac{\pi}{4}}\) log (1 + tan x) dx

⇒ I = \(\int_{0}^{\frac{\pi}{4}}\) log 2 dx – I [From Eq. (i)]
⇒ 2I = \([x \log 2]_{0}^{\frac{\pi}{4}}\)

⇒ 2I = \(\frac{\pi}{4}\) log 2

⇒ I = \(\frac{\pi}{8}\) log 2

Question 9.
\(\int_{0}^{2} x \sqrt{2-x}\) dx
Solution.
Let I = \(\int_{0}^{2} x \sqrt{2-x}\) dx

(∵ \(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx)

I = \(\int_{0}^{2}\) (2 – x) √x dx

Question 10.
\(\int_{0}^{\frac{\pi}{2}}\) (2 log sin x – log sin 2x) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}}\) (2 log sin x – log sin 2x) dx

⇒ I = \(\int_{0}^{\frac{\pi}{2}}\) {2 log sin x – log (2 sin x cos x)} dx

⇒ I = \(\int_{0}^{\frac{\pi}{2}}\) {2 log sin x – log sin x – log cos x – log 2} dx

⇒ I = \(\int_{0}^{\frac{\pi}{2}}\) {log sin x – log cos x – log 2}dx ……………(i)
We know that,
∵ \(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx
∴ I = \(\int_{0}^{\frac{\pi}{2}}\) {log cos x – log sin x – log 2} dx …………..(ii)
Adding equations (i) and (ii), we get
2I = \(\int_{0}^{\frac{\pi}{2}}\) (log 2 – log 2) dx

⇒ 2I = – 2 log 2 \(\int_{0}^{\frac{\pi}{2}}\) 1 dx

⇒ I = – log 2 [latex]\frac{\pi}{2}[/latex]

⇒ I = \(\frac{\pi}{2}\) (- log 2)

⇒ I = \(\frac{\pi}{2}\) [log \(\frac{1}{2}\)]

⇒ I = \(\frac{\pi}{2}\) log \(\frac{1}{2}\)

Question 11.
\(\int_{\frac{2}{2}}^{\frac{\pi}{2}}\) sin² x dx
Solution.
Let I = \(\int_{\frac{2}{2}}^{\frac{\pi}{2}}\) sin² x dx
As sin² (- x) = (sin(- x))²
= (- sin x)²
= sin 2x,
therefore, sin² x is an even function.
We know that if f(x)is an even function, then
\(\int_{-a}^{a}\) f(x)dx = 2 \(\int_{0}^{a}\) f(x) dx

I = 2 \(\int_{0}^{\frac{\pi}{2}}\) sin² x dx

= 2 \(\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2}\) dx

= \(\int_{0}^{\frac{\pi}{2}}\) (1 – cos 2x) dx

= \(\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\)

Question 12.
\(\int_{0}^{\pi} \frac{x d x}{1+\sin x}\)
Solution.

⇒ 2I = π \(\int_{0}^{\pi}\) {sec² x – tan x sec x} dx
⇒ 2I = π \([\tan x-\sec x]_{0}^{\pi}\)
⇒ 2I = π [(tan π – sec π) – (tan 0 – sec 0)
⇒ 2I = π [(0 – (- 1) – (0 – 1)]
⇒ 2I = π [2]
⇒ I = π

Question 13.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) sin⁷ x dx
Solution.
Let I = \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) sin⁷ x dx

As sin⁷ (- x) = (sin(- x))⁷
= (- sin x)⁷
= – sin⁷ x,
therefore, sin² x is an odd function.
We know that if f(x) is an odd function, then \(\int_{-a}^{a}\) f(x) dx = 0

∴ I = \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) sin⁷ x dx = 0.

Question 14.
\(\int_{0}^{2 \pi}\) cos⁵ x dx
Solution.
\(\int_{0}^{2 \pi}\) cos⁵ x dx = 2 \(\int_{0}^{\pi}\) cos⁵ x dx

(∵ \(\int_{0}^{2 a}\) f(x) = 2 \(\int_{0}^{a}\) f(x), where f(2 – a))

= f(x); hence 2a = 2x

∴ cos⁵ (2x – x) = cos⁵ x
= 2 × 0 = 0
(∵ \(\int_{0}^{2 a}\) f(x) = 0, if f(2a – x) = – f(x); hence 2a = π
∴ cos^{5</sup (π – x) = – cos5 x)}

Question 15.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}\) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}\) dx …………(i)

Question 16.
\(\int_{0}^{\pi}\) log (1 + cos x) dx
Solution.
Let I = \(\int_{0}^{\pi}\) log (1 + cos x) dx …………….(i)

⇒ I = \(\int_{0}^{\pi}\) log (1 + cos (π – x) dx
(∵ \(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx)

⇒ I = \(\int_{0}^{\pi}\) log (1 – cos x) dx ………….(ii)

⇒ 2I = \(\int_{0}^{\pi}\) {log (1 + cos x) + log (1 – cos x)} dx

⇒ 2I = \(\int_{0}^{\pi}\) log (1 + cos² x) dx

⇒ 2I = \(\int_{0}^{\pi}\) log sin² x dx

⇒ 2I = 2 \(\int_{0}^{\pi}\) log sin x

⇒ I = \(\int_{0}^{\pi}\) log sin x dx …………..(iii)

sin (π – x) = sin x
∴ I = 2 \(\int_{0}^{\frac{\pi}{2}}\) log sin x dx ………………(iv)

⇒ I = 2 \(\int_{0}^{\frac{\pi}{2}}\) log sin (\(\frac{\pi}{2}\) – x) dx

= 2 \(\int_{0}^{\frac{\pi}{2}}\) log cos x dx ………….(v)

Adding equations (iv) and (v), we get
2I = 2 \(\int_{0}^{\frac{\pi}{2}\) (log sin x + log cos x) dx

I = \(\int_{0}^{\frac{\pi}{2}\) (log sin x + log cos x + log 2 – log 2) dx

I = \(\int_{0}^{\frac{\pi}{2}\) (log 2 sin x cos x – log 2) dx

I = \(\int_{0}^{\frac{\pi}{2}\) log sin 2x dx – \(\int_{0}^{\frac{\pi}{2}\) log 2 dx
Let 2x = t
⇒ 2 dx = dt
When x = 0, t = o and when x = \(\frac{\pi}{2}\), t = π

∴ I = log sin t dt – \(\frac{\pi}{2}\) log 2

I = \(\frac{1}{2}\) I – \(\frac{\pi}{2}\) log 2

\(\frac{I}{2}\) = – \(\frac{\pi}{2}\) log 2

⇒ I = – π log 2.

Question 17.
\(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\) dx
Solution.

Question 18.
\(\int_{0}^{4}\) |x – 1| dx
Solution.
Let I = \(\int_{0}^{4}\) |x – 1| dx
It can be seen that, (x – 1) ≤ 0 when 0 ≤ x ≤ 1 and (x – 1) ≥ 0 when 1 ≤ x ≤ 4.

∴ I = \(\int_{0}^{1}\) |x – 1| dx + \(\int_{1}^{4}\) |x – 1| dx

(∵ \(\int_{a}^{b}\) f(x) = \(\int_{a}^{c}\) f(x) + \(\int_{c}^{b}\) f(x)dx)

= \(\int_{0}^{1}\) – (x – 1) dx + \(\int_{1}^{4}\) (x – 1) dx

= \(\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}\)

= \(1-\frac{1}{2}+\frac{(4)^{2}}{2}-4-\frac{1}{2}+1\)

= \(1-\frac{1}{2}+8-4-\frac{1}{2}+1\) = 5.

Question 19.
Show that \(\int_{0}^{a}\) f(x) g(x) dx = 2 \(\int_{0}^{a}\) f(x)dx if f and g are defined as
f(x) = f(a – x) and g(x) = g(a – x) = 4
Solution.
Let I = \(\int_{0}^{a}\) f(x) g(x) dx ………(i)

I = \(\int_{0}^{a}\) (a – x) g(a – x) dx

(∵ \(\int_{0}^{a}\) f(x) dx = \(\int_{0}^{a}\) f(a – x) dx)

= I = \(\int_{0}^{a}\) f(x)g (a – x) dx ……..(ii)

Adding equations (i) and (ii), we get

2I = \(\int_{0}^{a}\) {f(x) g(x) + f(x) g(a – x)} dx
⇒ 2I = \(\int_{0}^{a}\) f(x) {g(x) + g(a – x)}dx
⇒ 2I = \(\int_{0}^{a}\) f(x) × 4 dx
[∵ g(x) + g(a – x) = 4 given]
⇒ I = 2 \(\int_{0}^{a}\) f(x) dx.

Direction (20 – 21): Choose the correct answer.

Question 20.
The value of \(\int_{\frac{\pi}{2}}^{\frac{\pi}{2}}\) (x³ + x cos x + tan⁵ x + 1) dx is
(A) 0
(B) 2
(C) π
(D) 1
Solution.

Question 21.
The value of \(\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)\) dx is
(A) 2
(B) \(\frac{3}{4}\)
(C) 0
(D) – 1
Solution.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.10 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10

Direction (1 – 8): Evaluate the integrals using substitution.

Question 1.
\(\int_{0}^{1} \frac{x}{x^{2}+1}\) dx
Solution.
\(\int_{0}^{1} \frac{x}{x^{2}+1}\) dx
Let x² + 1 = t
⇒ 2x dx = dt
When x = 0, t = 1 and when x = 1, t = 2
∴ \(\int_{0}^{1} \frac{x}{x^{2}+1}\) dx = \(\frac{1}{2} \int_{1}^{2} \frac{d t}{t}\)

= \(\frac{1}{2}[\log |t|]_{1}^{2}\)

= \(\frac{1}{2}\) [log 2 – log 1]

= \(\frac{1}{2}\) log 2.

Question 2.
\(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi\)
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi\)

Also let sin Φ = t
⇒ cos Φ dΦ = dt
when, Φ = 0, t = 0 and
when Φ = \(\frac{\pi}{2}\), t = 1
∴ I = \(\int_{0}^{1}\) √t (1 – t²)² dt

Question 3.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) dx
Solution.
Let I = \(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\) dx
Also, let x = tan θ
⇒ dx = sec² θ dθ
when x = 0, θ = 0 and when x = 1, θ = \(\frac{\pi}{4}\)
I = \(\int_{0}^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\) sec² θ dθ

= \(\int_{0}^{\frac{\pi}{4}}\) sin^-1 (sin 2θ) sec² θ dθ

= \(\int_{0}^{\frac{\pi}{4}}\) 2θ sec² θ dθ

= 2 \(\int_{0}^{\frac{\pi}{4}}\) θ sec² θ dθ

Taking e as first function and sec² θ as second function and integrating by parts, we get
I = \(2\left[\theta \int \sec ^{2} \theta d \theta-\int\left\{\left(\frac{d}{d x}(\theta)\right) \int \sec ^{2} \theta d \theta\right\} d \theta\right]_{0}^{\frac{\pi}{4}}\)

= \(2\left[\theta \tan \theta-\int \tan \theta d \theta\right]_{0}^{\frac{\pi}{4}}=2[\theta \tan \theta+\log \mid \cos \theta]_{0}^{\frac{\pi}{4}}\)

Question 4.
\(\int_{0}^{2} x \sqrt{x+2}\) (Put x + 2 = t²)
Solution.
Let I = \(\int_{0}^{2} x \sqrt{x+2}\) dx
Also let x + 2 = t²
⇒ dx = 2t dt
when x = 0, t = √2 and when x = 2, t = 2

Question 5.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x}\) dx
Solution.
Let I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x}\) dx
Also, let cos x = t
⇒ – sin x dx = dt
When x = 0, t = 1 and when x = \(\frac{\pi}{2}\), t = 0
⇒ \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x}\) dx = \(-\int_{1}^{0} \frac{d t}{1+t^{2}}=-\left[\tan ^{-1} t\right]_{1}^{0}\)

= – [tan 0 – tan 1]

= – [- \(\frac{\pi}{4}\)]

= \(\frac{\pi}{4}\)

Question 6.
\(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)
Solution.
Let I = \(\int_{0}^{2} \frac{d x}{x+4-x^{2}}\)

= \(\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}\)

= \(\int_{0}^{2} \frac{d x}{-\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right)}\)

= \(\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}\)

= \(\int_{0}^{2} \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}\)
Also, let x – \(\frac{1}{2}\) = t
⇒ dx = dt
When x = 0, t = – \(\frac{1}{2}\) and when x = 2,t = \(\frac{3}{2}\)

Question 7.
\(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)
Solution.
Let I = \(\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}\)

= \(\int_{-1}^{1} \frac{d x}{\left(x^{2}+2 x+1\right)+4}\)

= \(\int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}\)

Also, let x + 1 = t
⇒ dx = dt
When x = – 1, t = 0 and when x = 1, t = 2

Question 8.
\(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x}\) dx
Solution.
Let I = \(\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x}\) dx

Also, let 2x = t
⇒ 2 dx = dt
When x = 1, t = 2 and when x = 2, t = 4

Direction (9 – 10): Choose the correct answer.

Question 9.
The value of the integral \(\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}}\) dx is
(A) 6
(B) 0
(C) 3
(D) 4
Solution.

Hence, the correct answer is (A).

Question 10.
If f(x) = \(\int_{0}^{x}\) t sin t dt, then f'(x) is
(A) cos x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x
Sol.
f(x) = \(\int_{0}^{x}\) t sin t dt
Integrating by parts, we get
f(x) = t \(\int_{0}^{x}\) sin t dt – \(\int_{0}^{x}\) {(\(\frac{d}{d t}\) t) ∫ sin t dt} dt
= \(\left[t(-\cos t]_{0}^{x}\right.\) – \(\int_{0}^{x}\) (- cos t) dt
= \([-t \cos t+\sin t]_{0}^{x}\)
= – x cos x + sin x
⇒ f’(x)= – [{x (- sin x)} + cos x + cos x
= x sin x – cos x + cos x
= x Sin x
Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5

Direction (1 – 21): Integrate the rational functions.

Question 1.
\(\frac{x}{(x+1)(x+2)}\)
Solution.
∫ \(\frac{x}{(x+1)(x+2)}\) dx

Let \(\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\)

⇒ \(\frac{x}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\)

⇒ x = A (x + 2) + B (x + 1)
Equating the coefficients of x and constant term, we get
A + B = 1 ……………(i)
2A + B = 0 ……………(ii)
On solving Eqs. (i) and (ii), we get
A = – 1 and B = 2
∴ \(\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\)

⇒ \(\int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x\)

= – log |x + 1| + 2 log |x + 2| + C

= log (x + 2)² – log |x + 1| + C

= log \(\left|\frac{(x+2)^{2}}{(x+1)}\right|\) + C

Question 2.
\(\frac{1}{x^{2}-9}\)
Solution.
\(\int \frac{1}{x^{2}-9} d x=\int \frac{1}{x^{2}-3^{2}} d x\)

= \(\int \frac{1}{(x+3)(x-3)} d x\)

Let \(\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}\)

1 = A (x – 3) + B (x + 3)
Equating the coefficients of x and constant term, we get
A + B = 0 and – 3A + 3B = 1
On solving, we get
A = – \(\frac{1}{6}\) and B = \(\frac{1}{6}\)

Question 3.
\(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
Solution.
∫ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) dx

Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\)

3x – 1 = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) …………..(i)
Substituting x = 1, 2 and 3 respectively in equation (i), we get
A = 1, B = – 5 and C = 4
∴ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\)

⇒ \(\int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right\} d x\)

= log |x – 1| – 5 log |x – 2| + 4 log |x – 3| + C

Question 4.
\(\frac{x}{(x-1)(x-2)(x-3)}\)
Solution.

Question 5.
\(\frac{2 x}{x^{2}+3 x+2}\)

Solution.

Question 6.
\(\frac{1-x^{2}}{x(1-2 x)}\)
Solution.
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (1 – x²) by x (1 – 2x), we get
\(\frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)\)

Let \(\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{(1-2 x)}\)
⇒ (2 – x) = A (1 – 2x) + Bx ……….(i)

Substituting x = 0 and \(\frac{1}{2}\) in equation (i) we get
A = 2 and B = 3

Question 7.
\(\left(x^{2}+1\right)(x-1)\)
Solution.
∫ \(\left(x^{2}+1\right)(x-1)\) dx
Let \(\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}\)

x = (Ax + B) (x – 1) + C (x² + 1)
⇒ x = Ax² – Ax + Bx – B + C
Equating the coefficients of x², x and constant term, we get
A + C = O;
– A + B = 1;
– B + C = 0
On solving, we get
A = – \(\frac{1}{2}\), B = \(\frac{1}{2}\) and C = \(\frac{1}{2}\)
From equation (1), we get

Question 8.
\(\frac{x}{(x-1)^{2}(x+2)}\)
Solution.
∫ \(\frac{x}{(x-1)^{2}(x+2)}\) dx
Let \(\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)}\)
x = A (x – 1) (x + 2) + B (x + 2) + C (x – 1)²
Substituting x = 1, we get
B = \(\frac{1}{3}\)
Equating the coefficients of x² and constant term, we get
A + C = 0;
– 2A + 2B + C = 0
On solving, we get
A = \(\frac{2}{9}\) and C = – \(\frac{2}{9}\)

Question 9.
\(\frac{3 x+5}{x^{3}-x^{2}-x+1}\)
Solution.
∫ \(\frac{3 x+5}{x^{3}-x^{2}-x+1}\) dx = \(\int \frac{3 x+5}{\left(x^{2}-1\right)(x-1)} d x=\int \frac{3 x+5}{(x-1)(x+1)(x-1)} d x\)

= \(\int \frac{3 x+5}{(x-1)^{2}(x+1)}\) dx

Let \(\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}\)

3x + 5 = A (x – 1) (x + 1) + B (x + 1) + C (x – 1)²
3x + 5 = A (x² – 1) + B (x + 1) + C (x + 1 – 2x) …………….(i)
Substituting x = 1 in equation (i), we get
B = 4
Equating the coefficients of x² and x, we get
A + C = 0;
B – 2C = 3
On solving, we get
A = – \(\frac{1}{2}\) and C = \(\frac{1}{2}\)
∴ \(\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}\)

Question 10.
\(\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}\)
Solution.
∫ \(\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}\) dx = ∫ \(\frac{2 x-3}{(x+1)(x-1)(2 x+3)}\) dx

Let \(\frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)}\)

⇒ (2x – 3) = A (x – 1) (2x + 3) + B (x + 1)(2x + 3) + C (x + 1) (x – 1)
⇒ (2x – 3) = A (2x² + x – 3) + B (2x² + 5x + 3) + C (x² – 1)
⇒ (2x – 3) = (2A + 2B + C) x² + (A + 5B) x + (- 3A + 3B – C)
Equating the coefficients of x2 and x, we get

Question 11.
\(\frac{5 x}{(x+1)\left(x^{2}-4\right)}\)
Solution.
∫ \(\frac{5 x}{(x+1)\left(x^{2}-4\right)}\) dx = ∫ \(\frac{5 x}{(x+1)(x+2)(x-2)}\) dx

Let \(\frac{5 x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\)

5x = A (x + 2) (x – 2) + B (x + 1) (x – 2) + C (x + 1) (x + 2)
Substituting x = – 1, – 2, and 2 respectively in equation (i), we get

Question 12.
\(\)
Solution.
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (x³ + x + 1) by x² – 1, we get

\(\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}\)

Let \(\frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}\)

⇒ 2x + 1 = A (x – 1) + B (x + 1) ………………(i)
Substituting x = 1 and – 1 in equation (i), we get

Question 13.
\(\frac{2}{(1-x)\left(1+x^{2}\right)}\)
Solution.
∫ \(\frac{2}{(1-x)\left(1+x^{2}\right)}\) dx

Let \(\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{(1-x)}+\frac{B x+C}{\left(1+x^{2}\right)}\)

2 = A (1 + x²) + (Bx + C) (1 – x)
⇒ 2 = A + Ax² + Bx – Bx² + C – Cx
Equating the coefficient of x², x, and constant term, we get
A – B = 0;
B – C = 0;
A + C = 2
On solving, we get
A = 1; B = 1 and C = 1

Question 14.
\(\frac{3 x-1}{(x+2)^{2}}\)
Solution.
∫ \(\frac{3 x-1}{(x+2)^{2}}\) dx

Let \(\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}\)

⇒ 3x – 1 = A (x + 2) + B
Equating the coefficient of x and constant term, we get
A = 3; 2A + B = – 1
⇒ B = – 7
∴ \(\frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\)

∫ \(\frac{3 x-1}{(x+2)^{2}}\) dx = \(3 \int \frac{1}{(x+2)} d x-7 \int \frac{x}{(x+2)^{2}} d x\)

= 3 log |x + 2| – 7 \(\left(\frac{-1}{(x+2)}\right)\) + C

= 3 log |x + 2| + \(\frac{7}{(x+2)}\) + C

Question 15.
\(\frac{1}{x^{4}-1}\)
Solution.
∫ \(\frac{1}{x^{4}-1}\) dx = \(\int \frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)} d x=\int \frac{1}{(x+1)(x-1)\left(1+x^{2}\right)} d x\)

Let \(\frac{1}{(x+1)(x-1)(1+x)^{2}}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^{2}+1\right)}\)

1 = A (x – 1) (x² + 1) + B (x + 1) (x² + 1) + (Cx + D) (x – 1)
1 = A (x³ + x – x² – 1) + B (x³ + x + x² + 1) + Cx³ + Dx² – Cx – D
1 = (A + B + C) x³ + (- A + B + D) x² + (A + B – C) x + (- A + B – D)
Equating the coefficient of x³, x², x and constant term, we get
A + B + C = 0;
– A + B + D = 0;
A + B – C = 0;
– A + B – D = 1
On solving, we get

Question 16.
\(\frac{1}{x\left(x^{n}+1\right)}\)
Solution.
[Hint : Multiply numerator and denominator by x^{n – 1} and put xⁿ = t]
Solution.
\(\frac{1}{x\left(x^{n}+1\right)}\)
Multiplying numerator and denominator by x^{n – 1}, we get

\(\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}\)

Let xⁿ = t
⇒ x^{n – 1} dx = dt

Question 17.
\(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) [Hint: Put sin x = t]
Solution.
∫ \(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) dx
Let sin x = t
⇒ cos x dx = dt

Question 18.
\(\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}\)
Solution.
\(\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}\) = 1 – \(\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}\)

Let \(\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)}\)

4x² + 10 = (Ax + B) (x² + 4) + (Cx + D) (x² + 3)
4x² + 10 = Ax³ + 4Ax + Bx² + 4B + Cx³ + 3Cx + Dx² + 3D
4x² + 10 = (A + C) x³ +(B + D)x² + (4A + 3C) x + (4B + 3D)
Equating the coefficients of x³, x², x and constánt term, we get
A + C = 0;
B + D = 4;
4A + 3C = 0;
4B + 3D = 10
On solving, we get
A = 0, B = – 2, C = 0 and D = 6

Question 19.
\(\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}\)
Solution.
∫ \(\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}\) dx
Let x² = t
⇒ 2x dx = dt

Question 20.
\(\frac{1}{x\left(x^{4}-1\right)}\)
Solution.
\(\frac{1}{x\left(x^{4}-1\right)}\)

Multiplying numerator and denominator by x³, we get

\(\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)}\)

Question 21.
\(\frac{1}{\left(e^{x}-1\right)}\) [Hint: Put e^x = t]
Solution.
∫ \(\frac{1}{\left(e^{x}-1\right)}\) dx
Let e^x = t
⇒ e^x dx = dt

⇒ \(\int \frac{1}{e^{x}-1} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t\)

Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\)

⇒ 1 = A (t – 1) + t …………..(i)

Substituting t = 1 and t = 0 in equation (i), we get
A = – 1 and B = 1

∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\)

⇒ ∫ \(\frac{1}{t(t-1)}\) dt = log \(\left|\frac{t-1}{t}\right|\) + C

= log \(\left|\frac{e^{x}-1}{e^{x}}\right|\) + C

Direction (22 – 23) Choose the correct answer:

Question 22.
∫ \(\frac{x d x}{(x-1)(x-2)}\) equals
(A) log \(\left|\frac{(x-1)^{2}}{x-2}\right|\) + C

(B) log \(\left|\frac{(x-2)^{2}}{x-1}\right|\) + C

(C) log \(\left|\left(\frac{x-1}{x-2}\right)^{2}\right|\) + C

(D) log |(x – 1) (x – 2)| + C
Solution.
Let \(\frac{x}{(x-1)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}\)

⇒ x = A (x – 2) + B (x – 1) …………….(i)

Substituting x = 1 and 2 in equation (i), we get
A = – 1 and B = 2

Hence, the correct answer is (B).

Question 23.
∫ \(\frac{d x}{x\left(x^{2}+1\right)}\) equals

(A) log |x| – \(\frac{1}{2}\) log (x² + 1) + C

(B) log |x| + \(\frac{1}{2}\) log (x² + 1) + C

(C) – log |x| + \(\frac{1}{2}\) log (x² + 1) + C

(D) \(\frac{1}{2}\) log |x| + log (x² + 1) + C
Solution.
Let \(\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}\)

⇒ 1 = A(x² + 1) + (Bx + C) x
Equating the coefficients of x², x and constant term, we get
A + B = 0;
C = 0; A = 1
On solving, we get
A = 1; B = – 1 and C = 0
∴ \(\frac{1}{x\left(x^{2}+1\right)}=\frac{1}{x}+\frac{-x}{x^{2}+1}\)

⇒ \(\int \frac{1}{x\left(x^{2}+1\right)} d x=\int\left\{\frac{1}{x}-\frac{x}{x^{2}+1}\right\} d x\)

= log |x| – \(\frac{1}{2}\) log (x² + 1) + C

Hence, the correct answer is (A).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8

Direction (1 – 6): Evaluate the following definite integrals as limit of sums.

Question 1.
\(\int_{a}^{b}\) x dx
Solution.
We know that,
\(\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]

Question 2.
\(\int_{0}^{5}\) (x + 1)
Solution.
Let I = \([\int_{0}^{5}/latex] (x + 1) dx
We know that,
[latex]\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]

where h = \(\frac{b-a}{n}\)
Here, a = 0, b = 5 and f(x) = (x + 1)

⇒ h = \(\frac{5-0}{n}=\frac{5}{n}\)

∴ \([\int_{0}^{5}/latex] (x + 1) dx = [latex](5-0) \lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+\ldots+f\left((n-1) \frac{5}{n}\right)\right]\)

= \(5 \lim _{n \rightarrow \infty} \frac{1}{n}\left[1+\left(\frac{5}{n}+1\right)+\ldots+\left\{1+\left(\frac{5(n-1)}{n}\right)\right\}\right]\)

Question 3.
\(\int_{2}^{3}\) x² dx
Solution.
We know that,
\(\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]

where h = \(\frac{b-a}{n}\)
Here, a = 2, b – 3 and f(x) = x²
⇒ h = \(\frac{3-2}{n}=\frac{1}{n}\)
∴ \(\int_{2}^{3}\) x² dx

Question 4.
\(\int_{1}^{4}\) (x² – x) dx
Solution.
We know that,
\(\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]
where nh = b – a
Given, \(\int_{1}^{4}\) (x² – x) dx,
here a = 1, b = 4 and nh = 3 and
f(x) = x² – x = x (x – 1)
∴ \(\int_{1}^{4}\) (x² – x) dx = \(\lim _{h \rightarrow 0}\) h [f(1) + f(1 + h) + f (1 + 2h) +……… + f (1 + (n – 1) h)]

= \(\lim _{h \rightarrow 0}\) h [1(1 – 1) + (1 + h) h + (1 + 2h) (2h) + ……………. + {1 + (n – 1) h} {(n – 1) h}] [∵ f(x) = x (x – 1)]
f(1) = 1 (1 – 1);
f(1 + h) = (1 + h) (1 + h – 1) = (1 + h) h and so on]
= \(\lim _{h \rightarrow 0}\) h² [(1 + h) + 2 (1 + 2h) + 3 (1 + 3h) + ……… + (n – 1)(1 + (n – 1) h)]

= \(\lim _{h \rightarrow 0}\) h² [{1 + 2 + 3 + ………….. + (n – 1)} + {h + 2² h + 3² h + …………. +(n – 1)²h)}]

Question 5.
\(\int_{-1}^{1}\) e^x dx
Solution.
Let I = \(\int_{-1}^{1}\) e^x dx ………… (i)
We know that,
\(\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]
where h = \(\frac{b-a}{n}\)
Here, a = – 1, = 1 and f(x) = e^x

Question 6.
\(\int_{0}^{4}\) (x + e^2x)dx
Solution.
We know that,
\(\int_{a}^{b}\) f(x) dx = (b – a) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(a) + f(a + h) + ……… + f(a + (n – 1) h]
where h = \(\frac{b-a}{n}\)
Here, a = 0, b = 4 and f(x) = x + e^2x
∴ h = \(\frac{4-0}{n}=\frac{4}{n}\)
⇒ \(\int_{0}^{4}\) (x + e^2x)dx = (4 – 0) \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [f(0) + f(h) + f(2h) + ……………. + f(n – 1) h)]

= 4 \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [(0 + e⁰) + (h + e^2h) + (2h + e^{2 . 2h}) + ……… {(n – 1) h + e^{2 (n – 1) h}}]

= 4 \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [1 + (h + e^2h) + (2h + e^4h) + …………. + (n – 1) h + e^{2 (n – 1) h}}] [∵ e⁰ = 1]

= 4 \(\lim _{n \rightarrow \infty} \frac{1}{n}\) [{h + 2h + 3h + ……….. + (n – 1) h} + (1 + e^2h + e^4h + ………….. + e^{2 (n – 1) h})}]

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7

Direction (1 – 9): Integrate the function.

Question 1.
\(\sqrt{4-x^{2}}\)
Solution.

Question 2.
\(\sqrt{1-4 x^{2}}\)
Solution.
Let I = ∫ \(\sqrt{1-4 x^{2}}\) dx

= ∫ \(\sqrt{(1)^{2}-(2 x)^{2}}\) dx

Let 2x = t
⇒ 2 dx = dt

Question 3.
\(\sqrt{x^{2}+4 x+6}\)
Solution.

Question 4.
\(\sqrt{x^{2}+4 x+1}\)
Solution.

Question 5.
\(\sqrt{1-4 x-x^{2}}\)
Solution.

Question 6.
\(\sqrt{x^{2}+4 x-5}\)
Solution.

Question 7.
\(\sqrt{1+3 x-x^{2}}\)
Solution.

Question 8.
\(\sqrt{x^{2}+3 x}\)
Solution.

Question 9.
\(\sqrt{1+\frac{x^{2}}{9}}\)
Solution.

Question 10.
∫ \(\sqrt{1+x^{2}}\) dx is equal to
(A) \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C

(B) \(\frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}\) + C

(C) \(\frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}\) + C

(D) \(\frac{x^{2}}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C
Solution.
We know that, ∫ \(\sqrt{a^{2}+x^{2}}\) dx = \(\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|\) + C

Here, a² = 1

∴ ∫ \(\sqrt{1+x^{2}}\) dx = \(\frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|x+\sqrt{1+x^{2}}\right|\) + C

Question 11.
∫ \(\sqrt{x^{2}-8 x+7}\) is equal to
(A) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C

(B) \(\frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|\) + C

(C) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C

(D) \(\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}+\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|\) + C
Solution.

Hence, the correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 1.
\(\frac{3 x^{2}}{x^{6}+1}\)
Solution.
∫ \(\frac{3 x^{2}}{x^{6}+1}\) dx = ∫ \(\frac{3 x^{2}}{\left(x^{3}\right)^{2}+1}\) dx
Let x³ = t
⇒ 3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^{2}}\)
∫ \(\frac{3 x^{2}}{\left(x^{3}\right)^{2}+1}\) dx = ∫ \(\frac{d t}{t^{2}+1}\)
= tan^-1 t + C
= tan^-1 (x³) + C

Question 2.
\(\frac{1}{\sqrt{1+4 x^{2}}}\)
Solution.
∫ \(\frac{1}{\sqrt{1+4 x^{2}}}\) dx = ∫ \(\) dx
Let 2x = t
⇒ 2 dx = dt
⇒ dx = \(\frac{1}{2}\)
∴ \(\int \frac{1}{\sqrt{1+4 x^{2}}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{1+t^{2}}}\)

Question 3.
\(\frac{1}{\sqrt{(2-x)^{2}+1}}\)
Solution.

Question 4.
\(\frac{1}{\sqrt{9-25 x^{2}}}\)
Solution.

Question 5.
\(\frac{3 x}{1+2 x^{4}}\)
Solution.
∫ \(\frac{3 x}{1+2 x^{4}}\) dx = \(\frac{3}{2} \int \frac{x d x}{\frac{1}{2}+x^{4}}=\frac{3}{2} \int \frac{x d x}{\frac{1}{2}+\left(x^{2}\right)^{2}}\)

Let x² = t
⇒ 2x = \(\frac{d t}{d x}\)

⇒ dx = \(\frac{d t}{2 x}\)

Question 6.
\(\frac{x^{2}}{1-x^{6}}\)
Solution.

Question 7.
\(\frac{x-1}{\sqrt{x^{2}-1}}\)
Solution.

Question 8.
\(\frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\)
Solution.

Question 9.
\(\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}\)
Solution.

Question 10.
\(\frac{1}{\sqrt{x^{2}+2 x+2}}\)
Solution.

Question 11.
\(\frac{1}{9 x^{2}+6 x+5}\)
Solution.

Question 12.
\(\frac{1}{\sqrt{7-6 x-x^{2}}}\)
Solution.
7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9).
Therefore, 7 – (x² + 6x + 9 – 9) = 16 – (x² + 6x + 9)
= 16 – (x + 3)²
= (4)² – (x + 3)²

Question 13.
\(\frac{1}{\sqrt{(x-1)(x-2)}}\)
Solution.
(x – 1) (x – 2) can be written as x² – 3x + 2

Question 14.
\(\frac{1}{\sqrt{8+3 x-x^{2}}}\)
Solution.
8 + 3x – x² can be written as – (x² – 3x + \(\frac{9}{4}\) – \(\frac{9}{4}\))

Therefore, 8 – (x² – 3x + \(\frac{9}{4}\) – \(\frac{9}{4}\)) = \(\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}\)

Question 15.
\(\frac{1}{\sqrt{(x-a)(x-b)}}\)
Solution.
(x – a) (x – b) can be written as x² – (a + b) x + ab
Therefore,

Question 16.
\(\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}\)
Solution.
∫ \(\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}\) dx
Let 2x² + x – 3 = t
⇒ (4x + 1) dx = dt
⇒ dx = \(\frac{d t}{4 x+1}\)

= \(\int \frac{d t}{\sqrt{t}}=\int t^{-\frac{1}{2}} d t=\frac{t^{-\frac{1}{2}+1}}{\frac{1}{2}}+C=2 \sqrt{t}+C\)

= 2 \(\sqrt{2 x^{2}+x-3}\) + C

Question 17.
\(\frac{x+2}{\sqrt{x^{2}-1}}\)

Solution.

Question 18.
\(\frac{5 x-2}{1+2 x+3 x^{2}}\)
Solution.
Let 5x – 2 = A \(\frac{d}{d x}\) (1 +2x + 3x²) + B
⇒ 5x – 2 = A (2 + 6x) + B
Equating the coefficient of x and constant term on both sides, we get
5 = 6A
⇒ A = \(\frac{5}{6}\) ;
2A + B = – 2
⇒ B = – \(\frac{11}{3}\)

Question 19.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}\)
Solution.
\(\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^{2}-9 x+20}}\)

Let 6x + 7 = A \(\frac{d}{d x}\) (x² – 9x + 20) + B
⇒ 6x + 7 = A (2x – 9) + B
Equating the coefficients of x and constant term, we get
2A = 6
⇒ A = 3;
– 9A + B = 7
⇒ B = 34
∴ 6x + 7 = 3 (2x – 9) + 34

Question 20.
\(\frac{x+2}{\sqrt{4 x-x^{2}}}\)
Solution.
∫ \(\frac{x+2}{\sqrt{4 x-x^{2}}}\) dx

Question 21.
\(\frac{x+2}{\sqrt{x^{2}+2 x+3}}\)
Solution.

Substituting equations (ii) and (iii) in equation (i), we get
∫ \(\frac{x+2}{\sqrt{x^{2}+2 x+3}}\) dx = \(\frac{1}{2}\left[2 \sqrt{x^{2}+2 x+3}\right]+\log \mid(x+1)+\sqrt{x^{2}+2 x+3 \mid}\)

= \(\sqrt{x^{2}+2 x+3}+\log \mid(x+1)+\sqrt{x^{2}+2 x+3 \mid}\) + C

Question 22.
\(\frac{x+3}{x^{2}-2 x-5}\)
Solution.
Let (x + 3) = A \(\frac{d}{d x}\) (x² – 2x – 5) + B
⇒ (x + 3) = A (2x – 2) + B
⇒ x + 3 = 2Ax – 2A + B
On equating the coefficients of x and constant term on both sides, we get
2A = 1
⇒ A = \(\frac{1}{2}\);
– 2A + B = 3
⇒ B = 4
⇒ (x + 3) = \(\frac{1}{2}\) (2x – 2) + 4

Question 23.
\(\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}\)
Solution.
Let 5x + 3 = A \(\frac{d}{d x}\) (x² + 4x + 10) + B
⇒ 5x + 3 = A (2x + 4) + B
⇒ 5x + 3 = 2Ax + 4A + B
On equating the coefficients of x and constant term, we get
2A = 5
A = \(\frac{5}{2}\);
4A + B = 3
⇒ B = – 7
⇒ 5x + 3 = (2x + 4) – 7

Direction (24 – 25): Choose the correct answer:

Question 24.
∫ \(\int \frac{d x}{x^{2}+2 x+2}\) equals
(A) x tan^-1 (x + 1) + C
(B) tan^-1 (x + 1) + C
(C) (x + 1) tan^-1 x + C
(D) tan^-1 x + C
Solution.
∫ \(\int \frac{d x}{x^{2}+2 x+2}\) dx = \(\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}=\int \frac{1}{(x+1)^{2}+(1)^{2}} d x\)
Let x + 1 = t
⇒ dx = dt
∴ ∫ \(\frac{1}{t^{2}+1^{2}}\) dt = \(\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)\) + C

= tan \(\left(\frac{x+1}{1}\right)\) + C

= tan^-1 (x + 1) + C
Hence, the correct answer is (B).

Question 25.
∫ \(\frac{d x}{\sqrt{9 x-4 x^{2}}}\) equals
(A) \(\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)\) + C

(B) \(\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)\)

(C) \(\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)\)

(D) \(\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)\)
Solution.

Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

Direction (1 – 22): Find the integrals of the functions:

Question 1.
sin² (2x + 5)
Solution.
= ∫ sin² (2x + 5) dx = ∫ \(\frac{1-\cos (4 x+10)}{2}\) dx
= ∫ 1 dx + \(\frac{1}{2}\) ∫ cos (4x + 10) dx
= \(\frac{1}{2}\) x – \(\frac{1}{2}\) (\(\left(\frac{\sin (4 x+10)}{4}\right)\)) + C
= \(\frac{x}{2}\) – \(\frac{1}{8}\) sin(4x + 10) + C

Question 2.
sin 3x cos 4x
Solution.
∫ sin 3x cos 4x dx = \(\frac{1}{2}\) ∫ {sin(3x + 4x) + sin(3x – 4x)} dx
[∵ 2 sin A cos B = sin (A + B) + sin(A – B)]
= \(\frac{1}{2}\) ∫ {sin 7x + sin (- x)}dx

= \(\frac{1}{2}\) ∫ {sin 7x – sin x} dx

= \(\frac{1}{2}\) ∫ sin 7x dx – \(\frac{1}{2}\) ∫ sin x dx

= \(\frac{1}{2}\) \(\left(\frac{-\cos 7 x}{7}\right)\) – \(\frac{1}{2}\) (- cos x) + C

= \(\frac{-\cos 7 x}{14}+\frac{\cos x}{2}\) + C

Question 3.
cos 2x cos 4x cos 6x
Solution.
∫ cos 2x (cos 4x cos 6x) dx = ∫ cos 2x [\(\frac{1}{2}\) {cos (4x +6x) + cos (4x – 6x)}] dx
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos 2x cos (- 2x)}dx 2J

= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos² 2x}dx

= \(\frac{1}{2}\) ∫ (cos 12x + cos 8x + 1 + cos 4x)dx

= \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C

Question 4.
sin³ (2x + 1)
Solution.
Let I = ∫ sin³ (2x + 1) dx
⇒ = ∫ sin² (2x + 1) . sin (2x + 1) dx
= ∫ (1 – cos² (2x + 1)) sin(2x +1) dx
Let cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
⇒ sin (2x + 1) dx = \(\frac{-d t}{2}\)
∴ I = – \(\frac{1}{2}\) ∫ (1 – t²) dt
= \(\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}\) + C

Question 5.
sin³ x cos³ x
Solution.
Let I = ∫ sin³ x cos³ x dx
= ∫ cos³ x . sin² x . sin x dx
= ∫ cos³ x (1 – cos² x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
⇒ I = – ∫ t³ (1 – t²) dt
= – ∫ (t³ – t⁵) dt
= – \(\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}\) + C

= – \(\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}\) + C

= \(\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}\) + C

Question 6.
sin x sin 2x sin 3x
Solution.
∫ sin x sin 2x sin 3x dx = ∫ [sin x . \(\frac{1}{2}\) {cos(2x – 3x) – cos(2x + 3x)}] dx
[∵ 2 sin A sin B = cos(A – B) – cos(A + B)]
= \(\frac{1}{2}\) ∫ (sin x cos (- x) – sin x cos 5x) dx
= \(\frac{1}{2}\) ∫ (sin x cos x – sin x cos 5x) dx

Question 7.
sin 4x sin 8x
Solution.
∫ sin 4x sin 8x dx = ∫ {\(\frac{1}{2}\) cos(4x – 8x) – cos(4x + 8x)} dx
= \(\frac{1}{2}\) ∫ (cos (- 4x) – cos 12x) dx
= \(\frac{1}{2}\) ∫(cos 4x – cos 12x) dx
= \(\frac{1}{2}\) \(\) + C
= \(\) [sin 4x – \([latex]\)[/latex] sin 12x] + C

Question 8.
\(\frac{1-\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{1-\cos x}{1+\cos x}\) dx = ∫ \(\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx

[∵ 1 – cos x = 2 sin² \(\frac{x}{2}\) and
1 + cos x = 2 cos² \(\frac{x}{2}\)]

= ∫ tan² \(\frac{x}{2}\) dx

= (sec² \(\frac{x}{2}\) – 1) dx

= ∫ sec² \(\frac{x}{2}\) dx – ∫ 1 dx

= \(\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]\) + C

= 2 tan \(\frac{x}{2}\) – x + C

Question 9.
\(\frac{\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{\cos x}{1+\cos x}\) dx = ∫ \(\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx
[∵ cos x = cos² \(\frac{x}{2}\) – sin² \(\frac{x}{2}\) and
cos x = 2 cos² \(\frac{x}{x}\) – 1]

= \(\frac{1}{2}\) ∫ (1 – tan² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ (1 – sec² \(\frac{x}{2}\) + 1 ) dx

= \(\frac{1}{2}\) ∫ (2 – sec² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ 2 dx – \(\frac{1}{2}\) ∫ sec² \(\frac{x}{2}\) dx

= \(\frac{1}{2}\) [2x – \(\frac{\tan \frac{x}{2}}{\frac{1}{2}}\)] + C

= x – tan \(\frac{x}{2}\) + C

Question 10.
sin⁴ x
Solution.
∫ sin⁴ x dx = ∫ sin² x sin² x dx

Question 11.
cos⁴ 2x
Solution.
∫ cos⁴ 2x dx = ∫ (cos² 2x)² dx

Question 12.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin ^{2} x}{1+\cos x}\) dx = ∫ \(\frac{\left(1-\cos ^{2} x\right)}{1+\cos x}\) dx

= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx

[∵ a² – ² = (a – b) (a + b)]

= ∫ (1 – cos x) dx = x – sin x + C

Question 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution.
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2\left(\cos ^{2} x-\cos ^{2} \alpha\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}\) dx

= 2 ∫(cos x + cos α) dx
= 2 sin x + 2 cos α x + C
= 2 (sin x + x cos α) + C

Question 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution.
∫ \(\frac{\cos x-\sin x}{1+\sin 2 x}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

Question 15.
tan³ 2x sec 2x
Solution.
∫ tan³ 2x sec 2x dx = ∫ tan² 2x tan 2x sec 2x dx
= ∫ (sec² 2x – 1) tan 2x sec 2x dx
= ∫ (sec² 2x tan 2x sec 2x – tan 2x sec 2x) dx
= ∫ sec² 2x tan 2x sec 2x dx – ∫ tan 2xsec 2x dx
= ∫ sec² 2x tan 2x sec 2x dx – \(\frac{\sec 2 x}{2}\) + C
Let sec 2x = t
∴ 2 sec 2x tan 2x dx = dt
∴ ∫ tan³ 2x sec 2x dx = \(\frac{1}{2}\) ∫ t² dt – \(\frac{\sec 2 x}{2}\) + C
= \(\frac{t^{3}}{6}-\frac{\sec 2 x}{2}\) + C

= \(\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}\) + C

Question 16.
tan4 x
Sol.
tan⁴ x dx = ∫ tan² x . tan² x dx
= ∫ [(sec² x – 1) tan² x] dx
= ∫ [sec² x tan² x – tan² x] dx
= ∫ sec² x tan² x dx – ∫ (sec² x – 1)dx
= ∫ sec² x tan² x dx – ∫ sec² x dx + ∫ 1 dx
= ∫ sec² x tan² x dx – tan x + x + C ……………….(i)
= ∫ sec² x tan² x dx
Let tan x = t
⇒ sec² x dx = dt
∫ sec² x tan² x dx = ∫ t² dt
= \(\frac{\tan ^{3} x}{3}\)
From Eq. (i), we get
∫ tan⁴ x dx = \(\frac{1}{3}\) tan³ x – tan x + x + C

Question 17.
\(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\)
Solution.
∫ \(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\) dx = \(\int \frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x\)

= \(\int \frac{\sin x}{\cos ^{2} x} d x+\int \frac{\cos x}{\sin ^{2} x} d x\)

= ∫ tan x sec x dx + ∫ cot x cosec x dx

= sec x – cosec x + C

Question 18.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Solution.
∫ \(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\) dx = ∫ \(\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x}\) dx
[∵ cos 2x = 1 – 2 sin² x]

∫ \(\frac{1}{\cos ^{2} x}\) dx
= ∫ sec² x dx = tan x + C

Question 19.
\(\frac{1}{\sin x \cos ^{3} x}\)
Solution.
∫ \(\frac{1}{\sin x \cos ^{3} x}\) dx

= ∫ tan x sec² x dx + ∫ \(\frac{\sec ^{2} x}{\tan x}\) dx
Let tan x = t
⇒ sec² x dx = dt
= ∫ t dt + ∫ \(\frac{1}{t}\) dt
= \(\frac{t^{2}}{2}\) + log |t| + C
= \(\frac{1}{2}\) tan² x + log |tan x| + C

Question 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\)
Solution.
∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = ∫ \(\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}\) dx
= ∫ \(\frac{\cos 2 x}{(1+\sin 2 x)}\) dx
Let 1 + sin 2x = t
⇒ 2 cos 2x dx = dt
∴ ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log|1 + sin 2x| + C

= \(\frac{1}{2}\) log |(sin x + cos x)²| + C

= log |sin x + cos x| + C

Question 21.
sin^-1 x (cos x)
Solution.
∫ sin^-1 x (cos x) dx = ∫ sin^-1 x [sin (\(\frac{\pi}{2}\) – x)] dx
= ∫ (\(\frac{\pi}{2}\) – x) = \(\frac{\pi}{2}\) ∫ dx – ∫ x dx
= \(\frac{\pi x}{2}-\frac{x^{2}}{2}\) + C

Question 22.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution.

Question 23.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx is equal to
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) tan x + cot x + C
(D) tan x + sec x + C
Solution.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx = ∫ \(\left(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right)\) dx
= ∫ (sec² x – cosec² x) dx
= tan x + cot x + C
Hence, the correct answer is (A).

Question 25.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx equals
(A) – cot (e x^x) + C
(B) tan (x e^x) + C
(C) tan (e^x) + C
(D) cot (e^x) + C
Solution.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx
Let x e^x = t
⇒ (e^x . x + e^x . 1) dx = dt
⇒ e^x (x + 1) dx = dt
∴ \(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x=\int \frac{d t}{\cos ^{2} t}\) = ∫ sec² t dt
= tan t + C
= tan (e^x x) + C
Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2

Direction (1 – 37): Integrate the following functions.

Question 1.
\(\frac{2 x}{1+x^{2}}\)
Solution.
∫ \(\frac{2 x}{1+x^{2}}\) dx
Let 1 + x² = t
On differentiating w.rt. x, we get
2x dx = dt
⇒ x dx = \(\frac{d t}{2}\)
⇒ ∫ \(\frac{2 x}{1+x^{2}}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + x²| + C

Question 2.
\(\frac{(\log x)^{2}}{x}\)
Sol.
∫ \(\frac{(\log x)^{2}}{x}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{(\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(\log x)^{3}}{3}\) + C

Question 3.
\(\frac{1}{x+x \log x}\)
Solution.
∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{x(1+\log x)}\) dx
Let 1 + log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
⇒ ∫ \(\frac{1}{x+x \log x}\) dx = ∫ \(\frac{1}{t}\) dt
= log |t| + C
= log |1 + log x| + C

Question 4.
sin x . sin (cos x)
Solution.
∫ sin x . sin(cos x) dx
Let cos x = t
On differentiating w.r.t. x, we get
– sin x dx = dt
⇒ ∫ sin x . sin (cos x) dx = – ∫ sin t dt
= – [- cos t] + C
= cos t + C = cos (cos x) + C

Question 5.
sin(ax + b) cos(ax + b)
Solution.
∫ sin(ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ 2 sin (ax + b) cos(ax + b) dx
= – \(\frac{1}{2}\) ∫ sin (2ax + 2b) dx
Let 2ax + 2b = t
On differentiating w.r.t. x, we get
2a dx = dt
∴ I = \(\frac{1}{2}\) ∫ sin t . \(\frac{d t}{2 a}\)

= \(\frac{1}{4 a}\) ∫ sin t dt

= – \(\frac{1}{4 a}\) cos t + C

= – \(\frac{1}{4 a}\) cos (2ax + 2) + C

Question 6.
\(\sqrt{a x+b}\)
Solution.
∫ \(\sqrt{a x+b}\) dx
Let ax + b = t
On differentiating w.r.t. x, we get
a dx = dt
∴ dx = \(\frac{1}{a}\) dt
⇒ ∫ \((a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{2}{3 a}(a x+b)^{\frac{3}{2}}\) + C

Question 7.
x \(\sqrt{x+2}\)
Solution.
∫ x \(\sqrt{x+2}\) dx
Let (x + 2) = t
On differentiating w.r.t. x, we get
dx = dt

Question 8.
x \(\sqrt{1+2 x^{2}}\)
Solution.
∫ x \(\sqrt{1+2 x^{2}}\) dx
Let 1 + 2x² = t
On differentiating w.r.t. x, we get
4x dx = dt
∫ x \(\sqrt{1+2 x^{2}}\) dx = \(\int \frac{\sqrt{t} d t}{4}=\frac{1}{4} \int t^{\frac{1}{2}} d t\)

= \(\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+C\)

Question 9.
(4x + 2) \(\sqrt{x^{2}+x+1}\)
Solution.
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx
Let x² + 2x + 1 = t
On differentiating w.r.t. x, we get
(2x + 1) dx = dt
∫ (4x + 2) \(\sqrt{x^{2}+x+1}\) dx = ∫ 2√t dt = 2 ∫ √t dt
= 2 \(\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\) + C

= \(\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}\) + C

Question 10.
\(\frac{1}{x-\sqrt{x}}\)
Solution.
∫ \(\frac{1}{x-\sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x}(\sqrt{x}-1)}\) dx
Let (√x – 1) = t
On differentiating w.r.t. x, we get
∴ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2√x dt
⇒ ∫ \(\frac{1}{2 \sqrt{x}}\) dx = ∫ \(\frac{1}{\sqrt{x} t} 2 \sqrt{x} d t=\int \frac{2}{t} d t\)
= 2 log |t| + C
= 2 log |\(\sqrt{x}-1\)| + C

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}\) dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

Question 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
Solution.
∫ \(\frac{x}{\sqrt{x+4}}[latex] dx
Let x + 4 = t
On differentiating w.r.t. x, we get
dx = dt

= [latex]\frac{2}{3}(x+4)^{\frac{1}{2}}\) (x + 4 – 12) + C

= \(\frac{2}{3} \sqrt{x+4}\) (x – 8) + C

Question 12.
\(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\)
Solution.
∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx
Let x³ – 1 = t
On differentiating w.r.t. x, we get
3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^{2}}\)
⇒ ∫ \(\left(x^{3}-1\right)^{\frac{1}{3}} x^{5}\) dx = ∫ (x³ – 1)^{\(\frac{4}{4}\)} x³ x² dx

Question 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
Solution.
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx
On differentiating w.r.t. x, we get
9x² dx = dt
⇒ dx = \(\frac{d t}{9 x^{2}}\)
∫ \(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\) dx = \(\frac{1}{9} \int \frac{d t}{(t)^{3}}=\frac{1}{9}\left[\frac{t^{-3+1}}{-3+1}\right]+C\)

= – \(\frac{1}{18}\left(\frac{1}{t^{2}}\right)\) + C

= – \(\frac{1}{18\left(2+3 x^{3}\right)^{2}}\) + C

Question 14.
\(\frac{1}{x(\log x)^{m}}\), x > 0, m ≠ 1
Solution.
∫ \(\frac{1}{x(\log x)^{m}}\) dx
Let log x = t
On differentiating w.r.t. x, we get
\(\frac{1}{x}\) dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{1}{x(\log x)^{m}}\) dx = ∫ \(\frac{d t}{(t)^{m}}\)

= \(\left(\frac{t^{-m+1}}{-m+1}\right)\) + C

= \(\frac{(\log x)^{1-m}}{(1-m)}\) + C

Question 15.
\(\frac{x}{9-4 x^{2}}\)
Solution.
∫ \(\frac{x}{9-4 x^{2}}\) dx
Let 9 – 4x² = t
On differentiating w.r.t. x, we get
– 8x dx = dt
⇒ dx = \(\frac{d t}{-8 x}\)
∫ \(\frac{x}{9-4 x^{2}}\) dx = \(\frac{-1}{8}\) ∫ \(\frac{1}{t}\) dt
= \(\frac{-1}{8}\) log |t| + C
= \(\frac{-1}{8}\) log |9 – 4x²| + C

Question 16.
e^{2x +3}
Solution.
∫ e^{2x + 3} dx
Let 2x + 3 = t
On differentiating w.r.t. x, we get
2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
∴ ∫ e^{2x + 3} dx = \(\frac{1}{2}\) ∫ e^t dt
= \(\frac{1}{2}\) (e^t) + C
= \(\frac{1}{2}\) e^{(2x + 3)} + C

Question 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
Solution.
∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx
Let tan^-1 x = t
⇒ \(\frac{1}{1+x^{2}}\) dx = dt
⇒ dx = (1 + x²) dt
∴ ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\) dx = ∫ e^t dt
= e^t + C
= e^tan-1 x + C

Question 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution.

= log |t| + C
= log |e^x + e^-x| + C.

Question 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
Solution.
∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx
Let e^2x + e^-2x = t
⇒ (2 e^2x – 2 e^-2x) dx = dt
⇒ dx = \(\frac{d t}{2\left(e^{2 x}-e^{-2 x}\right)}\)

∫ \(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\) dx = ∫ \(\frac{d t}{2 t}\)

= \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |e^2x + e^-2x| + C.

Question 21.
tan² (2x – 3)
Solution.
∫ tan² (2x – 3) dx = ∫ sec² (2x – 3) dx – ∫ 1 dx
Let 2x – 3 = t
⇒ 2 dx = dt
⇒ dx = \(\frac{1}{2}\) dt
⇒ ∫ sec² (2x – 3) dx – ∫ 1 dx = \(\frac{1}{2}\) ∫ (sec² t dt – ∫ 1 dx
= \(\frac{1}{2}\) tan t – x + C
= \(\frac{1}{2}\) tan (2x – 3) – x + C

Question 22.
sec² (7 – 4x)
Solution.
∫ sec² (7 – 4x) dx
Let 7 – 4x = t
⇒ – 4 dx = dt
⇒ dx = \(\frac{d t}{-4}\)

∴ ∫ sec² (7 – 4x) dx = \(\frac{-1}{4}\) ∫ sec² t dt
= \(\frac{-1}{4}\) (tan t) + C
= \(\frac{-1}{4}\) tan (7 – 4x) + C

Question 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Solution.
∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Let sin^-1 x = t

Question 24.
\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution.
∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}\) dx

Let 3 cos x + 2 sin x = t

⇒ (- 3 sin x + 2 cos x) dx = dt

⇒ dx = \(\frac{d t}{2 \cos x-3 \sin x}\)

∴ ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx = ∫ \(\frac{d t}{2 t}\) + \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log |2 sin x + 3 cos x| + C

Question 25.
\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\)
Solution.
∫ \(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\) dx = ∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx
Let (1 – tan x) = t
⇒ – sec² x dx = dt
⇒ dx = \(\frac{d t}{-\sec ^{2} x}\)

∫ \(\frac{\sec ^{2} x}{(1-\tan x)^{2}}\) dx = ∫ \(\frac{-d t}{t^{2}}\)
= – ∫ t² dt

= – \(\left(\frac{-t^{-2+1}}{-2+1}\right)\) + C

= \(\frac{1}{t}+C=\frac{1}{(1-\tan x)}\) + C

Question 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution.
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
Let √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ dx = 2 √x dt
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = 2 ∫ cos t dt
= 2 sin t + C
= 2 sin √x + C

Question 27.
\(\sqrt{\sin 2 x}\) cos 2x
Solution.
∫ \(\sqrt{\sin 2 x}\) cos 2x dx
Let sin 2x = t
⇒ 2 cos 2x dx = dt
⇒ dx = \(\frac{d t}{2 \cos 2 x}\)
∴ ∫ \(\sqrt{\sin 2 x}\) cos 2x dx = \(\frac{1}{2}\) ∫ (t)^{\(\frac{1}{2}\)} dt

= \(\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)\)

= \(\frac{1}{3} t^{\frac{3}{2}}\) + C

= \(\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}\) + C

Question 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution.
∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx
Let 1 + sin x = t
⇒ cos x dx = dt
⇒ dx = \(\frac{d t}{\cos x}\)

∫ \(\frac{\cos x}{\sqrt{1+\sin x}}\) dx = \(\frac{d t}{\sqrt{t}}\)

= \(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\) + C

= 2√t + C

= 2 \({\sqrt{1+\sin x}}\) + C

Question 29.
cot x log sin x
Solution.
∫ cot x log sin x dx
Let log sin x = t .
⇒ \(\frac{1}{\sin x}\) . cos x dx = dt
⇒ cot x dx = dt
⇒ dx = \(\frac{d t}{\cot x}\)
∴ ∫ cot x log sin x dx = ∫ t dt
= \(\frac{t^{2}}{2}\) + C
= \(\frac{1}{2}\) (log sin x)² + C

Question 30.
\(\frac{\sin x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin x}{1+\cos x}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{1+\cos x}\) dx = ∫ – \(\frac{d t}{t}\)
= – log |t| + C
= – log |1 + cos x| + C
= log \(\frac{1}{|1+\cos x|}\) + C

Question 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
Solution.
∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx
Let 1 + cos x = t
⇒ – sin x dx = dt
⇒ dx = \(=\frac{d t}{-\sin x}\)
∴ ∫ \(\frac{\sin x}{(1+\cos x)^{2}}\) dx = ∫ – \(\frac{d t}{t^{2}}\)

= ∫ t^-2 dt

= \(\frac{-t^{-2+1}}{-2+1}\) + C

= \(\frac{1}{t}\) + C

= \(\frac{1}{1+\cos x}\) + C

Question 32.
\(\frac{1}{1+\cot x}\)
Solution.

Question 33.
\(\frac{1}{1-\tan x}\)
Solution.

Question 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution.
Let I = \(\frac{\sqrt{\tan x}}{\sin x \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x}\) dx

= ∫ \(\frac{\sqrt{\tan x}}{\tan x \cos ^{2} x}\) dx

= ∫ \(\frac{\sec ^{2} x}{\sqrt{\tan x}}\) dx

Let x = tan t
⇒ sec² x dx = dt
⇒ dx = \(\frac{d t}{\sec ^{2} x}\)

∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)

= ∫ t^{\(-\frac{1}{2}\)}

= 2√t + C

= 2√tan x + C

Question 35.
\(\frac{(1+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(1+\log x)^{2}}{x}\) dx
Let 1 + log x = t
⇒ \(\frac{1}{x}\)dx = dt
⇒ dx = x dt
∴ ∫ \(\frac{(1+\log x)^{2}}{x}\) dx = ∫ t² dt
= \(\frac{t^{3}}{3}\) + C
= \(\frac{(1+\log x)^{3}}{3}\) + C

Question 36.
\(\frac{(x+1)(x+\log x)^{2}}{x}\)
Solution.
∫ \(\frac{(x+1)(x+\log x)^{2}}{x}\) dx = ∫ \(\frac{x+1}{x}\) (x + log x)² dx

= ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx

Let x + log x = t
⇒ (1 + \(\frac{1}{x}\)) dx = dt

∴ ∫ (1 + \(\frac{1}{x}\)) (x + log x)² dx = ∫ t² dt

= \(\frac{t^{3}}{3}\) + C

= \(\frac{1}{3}\) (x + log x)³ + C

Question 37.
\(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\)
Solution.
Let I = ∫ \(\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}\) dx

Let tan^-1 x⁴ = t
⇒ \(\frac{1}{1+x^{8}}\) . 4x³ = \(\frac{d t}{d x}=\)

⇒ dx = \(\frac{\left(1+x^{8}\right)}{4 x^{3}}\) dt

∴ I = ∫ \(\frac{x^{3} \sin t}{\left(1+x^{8}\right)} \cdot \frac{1+x^{8} d t}{4 x^{3}}\)

= \(\frac{1}{4}\) ∫ sin t dt

= – \(\frac{1}{4}\) cos t + C

Direction (38 – 39) : Choose the correct answer.

Question 38.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx equals
(A) 10^x – x¹⁰ + C

(B) 10^x + x¹⁰ + C

(C) (10^x – x¹⁰)^-1 + C

(D) log (10^x + x¹⁰) + C
Solution.
∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx
Let x¹⁰ + 10^x = t
⇒ dx = \(\frac{d t}{10 x^{9}+10^{x} \log _{e} 10}\)
∴ ∫ \(\frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}}\) dx = ∫ \(\frac{dt}{t}\)
= log t + C
= log (10^x + x¹⁰) + C
Hence, the correct answer is (D).

Question 39.
∫ \(\frac{d x}{\sin ^{2} x \cos ^{2} x}\) ,
(A) tan x + cot x + C
(B) tan x – cot x + C
(C) tan x cot x + C
(D) tan x – cot 2x + C
Solution.

Hence, the correct answer is (B).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.1

Direcflon (1 – 5):
Find the anti-derivative (or integral) of the following by the method of inspection.

Question 1.
sin 2x
Solution.
The anti-derivative of sin 2x is a function of x whose derivative is sin 2x.
It is known that,
\(\frac{d}{d x}\) (cos 2x) = – 2 sin 2x
⇒ sin 2x = – \(\frac{1}{2} \frac{d}{d x}\) (cos 2x)
sin 2x = \(\frac{d}{d x}\) (- \(\frac{1}{2}\) cos 2x)
Therefore, the anti-derivative of sin 2x is – \(\frac{1}{2}\) cos 2x + C.

Question 2.
cos 3x
Solution.
The anti-derivative of cos 3x is a function of x whose derivative is cos 3x.
It is known that,
\(\frac{d}{d x}\) (sin 3x) = 3 cos 3x
⇒ cos 3x = \(\frac{1}{3}\) \(\frac{d}{d x}\) (sin 3x)

∴ cos 3x = \(\frac{d}{d x}\) (\(\frac{1}{3}\) sin 3x)

Therefore, the anti-derivative of cos 3x is \(\frac{1}{3}\) sin 3x + C.

Question 3.
e^2x
Solution.
The anti-derivative of e^2x is the function of x whose derivative is e^2x.
It is known that,
\(\frac{d}{d x}\) (e^2x) = 2e^2x
⇒ e^2x = \(\frac{1}{2}\) \(\frac{d}{d x}\) (e^2x)

∴ e^2x = \(\frac{d}{d x}\) (\(\frac{1}{2}\) e^2x)

Therefore, the anti-derivative of e^2x is \(\frac{1}{2}\) e^2x + C.

Question 4.
(ax + b)²
Solution.
The anti-derivative of (ax + b)² is the function of x whose derivative is (ax + b)².
It is known that,
\(\frac{d}{d x}\) (ax + b)² = 3a (ax + b)²

⇒ (ax + b)² = \(\frac{1}{3 a}\) \(\frac{d}{d x}\) (ax + b)³

∴ (ax + b)² = \(\frac{d}{d x}\) (\(\frac{1}{3 a}\) (ax + b)³))

Therefore, the anti-derivative of (ax + b)² is \(\frac{1}{3 a}\) (ax + b)³) + C.

Question 5.
sin 2x – 4 e^3x
Solution.
The anti-derivative of (sin 2x – 4 e^3x) is the function of x whose derivative is (sin 2x – 4 e^3x).
It is known that,
\(\frac{d}{d x}\) (- \(\frac{1}{2}\) cos 2x – \(\frac{4}{3}\) e^3x) = sin 2x – 4 e^3x
Therefore, the anti-derivative of (sin 2x – 4 e^3x) is (- \(\frac{1}{2}\) cos 2x – \(\frac{4}{3}\) e^3x) + C.

Direction (6 – 20): Find the following integrals:

Question 6.
∫ (4 e^3x + 1) dx
Solution.
∫ (4 e^3x + 1) dx = 4 ∫ e^3x dx + ∫ 1 dx

= 4 \(\left(\frac{e^{3 x}}{3}\right)\) + x + C

= \(\frac{4}{3}\) e^3x + x + C {∵ ∫ e^nx dx = \(\frac{e^{n x}}{x}\)}

Question 7.
∫ x² (1 – \(\frac{1}{x^{2}}\)) dx
Solution.
∫ x² (1 – \(\frac{1}{x^{2}}\)) dx = ∫ (x² – 1) dx
= ∫ x² dx – ∫ 1 dx
= \(\frac{x^{3}}{3}\) – x + C.

Q.8.
∫ (ax² + bx + c) dx
Solution.
∫ (ax² + bx + c) dx = a ∫ x² dx + b ∫ x dx + c ∫ 1 dx

= \(\frac{a x^{2+1}}{2+1}+\frac{b x^{1+1}}{1+1}\) + cx + C

= \(a\left(\frac{x^{3}}{3}\right)+b\left(\frac{x^{2}}{2}\right)\) + cx + C

= \(\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\) + cx + C

Question 9.
∫ (2x² + e^x) dx
Solution.
∫ (2x² + e^x) dx = 2 ∫ x² dx + ∫ e^x dx

= 2 \(\left(\frac{x^{2+1}}{2+1}\right)\) + e^x + C

= 2 \(\left(\frac{x^{3}}{3}\right)\) + e^x + C

= \(\frac{2}{3}\) x³ + e^x + C

Question 10.
∫ (√x – \(\frac{1}{\sqrt{x}}\))² dx
Solution.
∫ (√x – \(\frac{1}{\sqrt{x}}\))² dx = ∫ [(√x)² – 2 √x × \(\frac{1}{\sqrt{x}}\)] dx

= ∫ (x + \(\frac{1}{x}\) – 2) dx = ∫ x dx + ∫ \(\frac{1}{x}\) dx – 2 ∫ 1 dx

= \(\frac{x^{2}}{2}\) + log |x| – 2x + C.

Question 11.
∫ \(\frac{x^{3}+5 x^{2}-4}{x^{2}}\) dx
Solution.
∫ \(\frac{x^{3}+5 x^{2}-4}{x^{2}}\) dx = ∫ \(\left(\frac{x^{3}}{x^{2}}+5 \frac{x^{2}}{x^{2}}-\frac{4}{x^{2}}\right)\) dx

= ∫ (x + 5 – 4 x^-2) dx
= ∫ x dx + 5 ∫ 1 dx – 4 ∫ x^-2 dx
= \(\frac{x^{2}}{2}\) + 5x – 4 \(\left(\frac{x^{-2+1}}{-2+1}\right)\) + C

= \(\frac{x^{2}}{2}\) + 5x – 4 \(\left(\frac{x^{-1}}{-1}\right)\) + C

= \(\frac{x^{2}}{2}\) + 5x + \(\frac{4}{4}\) + C

Question 12.
∫ \(\frac{x^{3}+3 x+4}{\sqrt{x}}\) dx
Solution.
∫ \(\frac{x^{3}+3 x+4}{\sqrt{x}}\) dx

=

Question 13.
∫ \(\frac{x^{3}-x^{2}+x-1}{x-1}\) dx
Solution.
∫ \(\frac{x^{3}-x^{2}+x-1}{x-1}\) dx = ∫ \(\frac{x^{2}(x-1)+1(x-1)}{x-1}\) dx

= ∫ \(\frac{\left(x^{2}+1\right)(x-1)}{x-1}\) dx

= ∫ (x² + 1) dx

= ∫ x² dx + 1 dx

= \(\frac{x^{3}}{3}\) + x + C.

Question 14.
∫ (1 – x) √x dx
Solution.
∫ (1 – x) √x dx = ∫ \(\left(x^{\frac{1}{2}}-x^{\frac{3}{2}}\right)\) dx

=

Question 15.
∫ √x (3x² + 2x + 3) dx
Solution.
∫ √x (3x² + 2x + 3) dx

=

Question 16.
∫ (2x – 3 cos x + e^x) dx
Solution.
∫ (2x – 3 cos x + e^x) dx = 2 ∫ x dx – 3 ∫ cos x dx + ∫ e^x dx
= \(\frac{2 x^{2}}{2}\) – 3 (sin x) + e^x + C
= x² – 3 sin x + e^x + C

Question 17.
∫ (2x² – 3 sin x + 5√x) dx
Solution.
∫ (2x² – 3 sin x + 5√x) dx = 2 ∫ x² dx – 3 ∫ sin x dx + 5 ∫ x^{\(\frac{1}{2}\)} dx

= \(\frac{2 x^{3}}{3}\) – 3 (- cos x) + 5 \(\left(\frac{x^{3 / 2}}{\frac{3}{2}}\right)\) + C

= \(\frac{2}{3}\) x³ + 3 cos x + \(\frac{10}{3}\) x^{\(\frac{3}{2}\)} + C.

Question 18.
∫ sec x (sec x + tan x) dx
Solution.
∫ sec x (sec x + tan x) dx = ∫ (sec² x + sec x . tan x dx
= ∫ sec² x dx + ∫ sec x . tan x dx
= tan x + sec x + C

Question 19.
∫ \(\frac{\sec ^{2} x}{\ {cosec}^{2} x}\) dx
Solution.
∫ \(\frac{\sec ^{2} x}{\ {cosec}^{2} x}\) dx = ∫ \(\frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}}\) dx

= ∫ \(\frac{\sin ^{2} x}{\cos ^{2} x}\) dx

= ∫ tan² x dx

= ∫ (sec² x – 1) dx

= ∫ sec² x dx – ∫ 1 dx = tan x – x + C

Question 20.
∫ \(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\)
Solution.
∫ \(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\) = ∫ \(\left(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}\right)\) dx

= ∫ (2 sec² x – 3 sec x . tan x) dx
= 2 ∫ sec² x dx – ∫ tan x . sec x dx
= 2 tan x – 3 sec x + C.

Direction (21 – 22) : Choose the correct answer.

Question 21.
The anti-derivative of (√x + \(\frac{1}{\sqrt{x}}\)) equals.
(A) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)

(B) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+C\)

(C) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)

(D) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Solution.
∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx = ∫ \(x^{\frac{1}{2}}\) dx + ∫ \(x^{-\frac{1}{2}}\) dx

= \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\) + C

= \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}\) + C

Hence the correct answer is (C).

Question 22.
If \(\frac{d}{d x}\) f(x) = 4x³ – \(\frac{3}{x^{4}}\) such that f(2) = 0. Then, f(x) is
(A) x⁴ + \(\frac{1}{x^{3}}-\frac{129}{8}\)

(B) x⁴ + \(\frac{1}{x^{4}}+\frac{129}{8}\)

(C) x³ + \(\frac{1}{x^{3}}+\frac{129}{8}\)

(D) x³ + \(\frac{1}{x^{4}}-\frac{129}{8}\)
Solution.
It is given that,
\(\frac{d}{d x}\) f(x) = 4x³ – \(\frac{3}{x^{4}}\)

∴ Anti – derivative of 4x³ – \(\frac{3}{x^{4}}\) = f(x)

∴ f(x) = ∫ (4x³ – \(\frac{3}{x^{4}}\)) dx

⇒ f(x) = 4 ∫ x³ dx – 3 ∫ x^-4 dx

f(x) = \(4\left(\frac{x^{4}}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C\)

⇒ f(x) = x⁴ + \(\frac{1}{x^{3}}\) + C
Also, f(2) = 0
∴ f(2) = (2)⁴ + \(\frac{1}{(2)^{3}}\) + C

⇒ 16 + \(\frac{1}{8}\) + C = 0

⇒ C = – (16 + \(\frac{1}{8}\))

⇒ C = – \(\frac{129}{8}\)

⇒ f(x) = x⁴ + \(\frac{1}{x^{3}}-\frac{129}{8}\)

Hence, the correct answer is (A).