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.
e2x +3
Solution.
∫ e2x + 3 dx
Let 2x + 3 = t
On differentiating w.r.t. x, we get
2 dx = dt
⇒ dx = \frac{1}{2} dt
∴ ∫ e2x + 3 dx = \frac{1}{2} ∫ et dt
= \frac{1}{2} (et) + 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 = ∫ et dt
= et + C
= etan-1 x + C
Question 19.
\frac{e^{2 x}-1}{e^{2 x}+1}
Solution.
= log |t| + C
= log |ex + 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 e2x + e-2x = t
⇒ (2 e2x – 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 |e2x + 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) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)-1 + C
(D) log (10x + x10) + C
Solution.
∫ \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} dx
Let x10 + 10x = 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 (10x + 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).