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PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

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

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

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

Let \frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x} ………………..(i)

Equating the coefficients of x2, x, and constant term, we get
– A + B – C = 0
B + C = 0, A = 1
On solving these equations, we get
A = 1, B = \frac{1}{2} and C = – \frac{1}{2}
From equation (i), we get

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 1

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 2.
\frac{1}{\sqrt{x+a}+\sqrt{(x+b)}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 2

= \frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]

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

Question 3.
\frac{1}{x \sqrt{a x-x^{2}}} [Hint: Put x = \frac{a}{t}]
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 3

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

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

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 4

Question 5.
Evaluate: \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} [Hint: , put x = t6]
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 5

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 6.
\frac{5 x}{(x+1)\left(x^{2}+9\right)}
Solution.
Let \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)} ………..(i)
⇒ 5x = A(x2 + 9) + (Bx + C) (x + 1)
⇒ 5x = Ax2 + 9A + Bx2 +Bx + Cx + C
Equating the coefficients of x2, x, and constant term, we get
A + B = 0;
B + C = 5;
9A + C = 0
On solving these equations, we get
A = – \frac{1}{2}; B = \frac{1}{2} and C = \frac{9}{2}
From equation (i), we get

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 6

Question 7.
\frac{\sin x}{\sin (x-a)}
Solution.
\frac{\sin x}{\sin (x-a)}
Let x – a = t
⇒ dx = dt
\frac{\sin x}{\sin (x-a)} dx = ∫ \frac{\sin (t+a)}{\sin t} dt

= ∫ \frac{\sin t \cos a+\cos t \sin a}{\sin t} dt

= ∫ (cos a + cos t sin a) dt
= t cos a + sin a log |sin t| + C1
= (x – a) cos a + sin a log sin (x – a) + C1
= x cos a + sin a log sin (x – a) – a cos a + C1
= sin a log sin (x – a) + x cos a + C

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 8.
\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 7

Question 9.
\frac{\cos x}{\sqrt{4-\sin ^{2} x}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 8

Question 10.
\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 9

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 11.
\frac{1}{\cos (x+a) \cos (x+b)}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 10

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 11

Question 12.
\frac{x^{3}}{1-x^{8}}
Solution.
\frac{x^{3}}{1-x^{8}}
Let x4 = t
⇒ 4x3 dx = dt
\int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}

= \frac{1}{4} sin-1 t + C

= \frac{1}{4} sin-1 (x4) + C
[∵ ∫ \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)]

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

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

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 12

Question 14.
\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}
Solution.
\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)} …………..(i)

⇒ 1 = (Ax + B) (x2 + 4) + (Cx + D) (x2 + 1)
⇒ 1 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + Cx + Dx2 + D
Equating the coefficients of x3, x2, x, and constant term, we get
A + C = 0;
B + D = 0;
4A + C = 0;
4B + D = 1
On solving these equations, we get
A = 0, B = 1, C = 0 and D = – \frac{1}{3}

From equation (i), we get

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 13

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 15.
cos3 x elog sin x
Solution.
cos3 x elog sin x = cos3 × x sin x
Let cos x = t
⇒ – sin x dx = dt
⇒ ∫ cos3 x elog sin x dx = ∫ cos3 x sin x dx
= – ∫ t3 dt
= \frac{t^{4}}{4} + C

= – \frac{\cos ^{4} x}{4} + C

Question 16.
e3 log x (x4 + 1)-1
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 14

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 17.
f'(ax + b) [f(ax + b)]n
Solution.
f'(ax + b) [f(ax + b)]n
Let f (ax + b) = t
⇒ af’(ax + b) dx = dt
⇒ f’(ax + b)[f(ax + b)]n dx = \frac{1}{a} ∫ tn dt

= \frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]

= \frac{1}{a(n+1)} f(ax + b)n+1 + C

Question 18.
\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}
Solution.
\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}}

= \frac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}}

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 15

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 19.
\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x ∈ [0, 1]
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 16

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 17

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 20.
\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 18

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 19

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 21.
\frac{2+\sin 2 x}{1+\cos 2 x} ex
Solution.
Let I = ∫ (\frac{2+\sin 2 x}{1+\cos 2 x}) ex dx

= ∫ \left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right) ex dx

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

= ∫ (sec2 x + tan x) ex dx
Let f(x) = tan x
⇒ f’(x) = sec2 x
I = ∫ [f(x) + f'(x)] ex dx
= ex f(x) + C
= ex tan x + C

Question 22.
\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}
Solution.
\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)} …………..(i)

⇒ x2 + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C(x2 +2x+1)
⇒ x2 + x + 1 = A(x2 + 3x + 2) + B (x + 2) + C(x2 + 2x + 1)
⇒ x2 + x + 1 = (A + C) x2 + (3A +B + 2C) x + (2A + 2B + C)
Equating the coefficients of x2, x, and constant term, we get
A + C = 1;
3A + B + 2C = 1;
2A + 2B + C = 1
On solving these equations, we get
A = – 2, B = 1 andC = 3
From equation (i), we get

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

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

= – 2 log |x + 1| + 3 log |x + 2| – \frac{1}{(x+1)} + C.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 23.
tan-1 \sqrt{\frac{1-x}{1+x}}
Solution.
Let I = ∫ tan-1 \sqrt{\frac{1-x}{1+x}} dx
Let x = cos θ
⇒ dx = – sin θ dθ
I = ∫ tan-1 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (- sin θ dθ)

= – ∫ tan-1 \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} sin θ dθ

= – ∫ tan-1 tan \frac{\theta}{2} . sin θ dθ

= – \frac{1}{2} ∫ θ . sin θ dθ

= – \frac{1}{2} [θ – cos θ] – ∫ 1 . (- cos θ) dθ

= – \frac{1}{2} [- θ cos θ + sin θ]

= + \frac{1}{2} θ cos θ – \frac{1}{2} sin θ

= \frac{1}{2} cos-1 x . x – \frac{1}{2} \sqrt{1-x^{2}} + C

= \frac{x}{2} cos-1 x – \frac{1}{2} \sqrt{1-x^{2}} + C

= \frac{1}{2} [x cos-1 x – \sqrt{1-x^{2}}) + C

Question 24.
\frac{\sqrt{x^{2}+1\left[\log \left(x^{2}+1\right)-2 \log x\right]}}{x^{4}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 20

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 21

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Direction (25 – 33): Evaluate the definite integral.

Question 25.
\cdot \int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) dx
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 23

Question 26.
\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} dx
Solution.
Let I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} dx

= ∫ \int_{0}^{\frac{\pi}{4}} \frac{\cos ^{4} x}{\frac{\cos ^{4} x+\sin ^{4} x}{\cos ^{4} x}} dx

= ∫ \int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x} dx

Let tan2 x = t
⇒ 2 tan x sec2 x dx = dt
When x = 0, t = 0 and when x = \frac{\pi}{4}, t = 1
∴ I = \frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1}

= \frac{1}{2} [tan-1 1 – tan-21 0]

= \frac{1}{2}\left[\frac{\pi}{4}\right]=\frac{\pi}{8}

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 27.
\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x} dx
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 27

Let 2 tan x = t
⇒ 2 sec2 x dx = dt
When x = 0, t = 0 and when x = \frac{\pi}{2}, t = ∞
⇒ I = \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\int_{0}^{\infty} \frac{d t}{1+t^{2}}

= \left[\tan ^{-1} t\right]_{0}^{\infty}

= [tan (∞) – tan (o)] = \frac{\pi}{2}
Therefore, from Eq. Ci), we get
I = -\frac{\pi}{6}+\frac{2}{3}\left[\frac{\pi}{2}\right]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}

Question 28.
\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}
Solution.
Let I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} dx

= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}} dx

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 24

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 29.
\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 25

Question 30.
\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} dx
Solution.
Let I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} dx
Also, let sin x – cos x = t
⇒ (cos x + sin x) dx = dt
When x = 0, t= – 1 and when x =\frac{\pi}{4}, t = 0
⇒ (sin x – cosx)2 = t2
⇒ sin2 x + cos2 x – 2 sin x cos x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 26

Question 31.
\int_{0}^{\frac{\pi}{2}} sin 2x tan-1 (sin x) dx
Solution.
Let I = \int_{0}^{\frac{\pi}{2}} sin 2x tan-1 (sin x) dx

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

Let sin x = t
cos dx = dt
when x = 0 , t = 0 and when x = \frac{\pi}{2}, then t = 1

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 27

Question 32.
\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} dx
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 28

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 29

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 33.
\int_{1}^{4} [|x – 1| + |x – 2| + |x – 3|] dx
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 30

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 31

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Direction (34 – 39): Prove the following:

Question 34.
\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}
Solution.
Let I = \int_{1}^{3} \frac{d x}{x^{2}(x+1)}

Also, let \frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}

1 = Ax (x + 1) + B(x + 1) + C (x2)
1 = Ax2) + Ax + Bx + B + Cx2)
Equating the coefficients of x2), x and constant term, we get
A + C = 0;
A + B = 0;
B = 1
On solving these equations, we get
A = – 1; C = 1 and B = 1

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 32

Hence, the given result is proved.

Question 35.
\int_{0}^{1} x ex dx = 1
Solution.
Let I = \int_{0}^{1} x ex dx dx
Integrating by parts, we get
I = x \int_{0}^{1} e^{x} d x-\int_{0}^{1}\left\{\left(\frac{d}{d x}(x)\right) \int e^{x} d x\right\} dx

= \left[x e^{x}\right]_{0}^{1}\int_{0}^{1} ex dx

= \left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}

= e – e + 1 = 1

Hence, the given result is proved.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 36.
\int_{-1}^{1} x17 cos4 x dx
Solution.
Let I = \int_{-1}^{1} x17 cos4 x dx
Also, f(x) = (- x)17 cos4 (- x)
= – x17 cos4 x = – f(x)
Therefore, f(x) is an odd function.
We know that if f(x) is an odd function, then \int_{-a}^{a} f(x) dx = o
∴ I = \int_{-1}^{1} x17 cos4 x dx = o
Hence, the given result is proved.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 37.
\int_{0}^{\frac{\pi}{2}} sin2 x dx = \frac{2}{3}
Solution.
Let I = \int_{0}^{\frac{\pi}{2}} sin2 x dx

= \int_{0}^{\frac{\pi}{2}} sin3 x sin x dx

= \int_{0}^{\frac{\pi}{2}} (1 – cos2 x) sin x dx

= \int_{0}^{\frac{\pi}{2}} sin x dx – \int_{0}^{\frac{\pi}{2}} cos2 x sin x dx

= [-\cos ]_{0}^{\frac{\pi}{2}}+\left[\frac{\cos ^{3} x}{3}\right]_{0}^{\frac{\pi}{2}}

= 1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}
Hence, the given result is proved.

Question 38.
\int_{0}^{\frac{\pi}{4}} 2 tan3 x dx = 1 – log 2
Solution.
Let I = \int_{0}^{\frac{\pi}{4}} 2 tan3 x dx

= 2 \int_{0}^{\frac{\pi}{4}} tan2 x tan x dx

= 2 \int_{0}^{\frac{\pi}{4}} (sec2 x – 1) tan x dx

= 2 \int_{0}^{\frac{\pi}{4}} sec2 x tan x dx – 2 \int_{0}^{\frac{\pi}{4}} tan x dx
= 2\left[\frac{\tan ^{2} x}{2}\right]_{0}^{\frac{\pi}{4}}+2[\log \cos x]_{0}^{\frac{\pi}{4}}

=1 + 2 [log cos \frac{\pi}{4} – log cos 0]
= 1 + 2 [log \frac{1}{\sqrt{2}} – log 1]
= 1 – log 2 – log 1
= 1 – log 2
Hence, the given result is proved.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise

Question 39.
\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1
Solution.
Let I = \int_{0}^{1} sin-1 x dx
= \int_{0}^{1} sin-1 x . 1. dx

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 33

Hence, the given result is proved.

Question 40.
Eva1uate \int_{0}^{1} e2 – 3x dx as a limit of a sum.
Solution.
Let I = \int_{0}^{1} e2 – 3x dx
We know that,

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 34

Direction (41 – 44): Choose the correct answer:

Question 41.
\int \frac{d x}{e^{x}+e^{-x}} is equal to
(A) tan-1 (ex) + C
(B) tan-1 (ex) + C
(C) log (ex – e-x) + C
(D) log (ex + e-x) + C
Solution.
Let I = \int \frac{d x}{e^{x}+e^{-x}}
= ∫ \frac{e^{x}}{e^{2 x}+1} dx
Also, let ex = t
⇒ ex dx = dt
∴ I = ∫ \frac{d t}{1+t^{2}} tan-1 t + C
= tan-1 (ex) + C
Hence, the correct answer is (A).

Question 42.
\frac{\cos 2 x}{(\sin x+\cos x)^{2}} dx is equal to

(A) \frac{-1}{\sin x+\cos x}

(B) log |sin x + cos x| + C

(C) log |sin x – cos x| + C

(D) \frac{1}{(\sin x+\cos x)^{2}}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 35

Let cos x + sin x = t
⇒ (cos x – sin x) dx = dt
∴ I = \int \frac{d t}{t}
= log |t| + C
= log |cos x+ sin x| + C
Hence, the correct answer is (B).

Question 43.
If f(a + b – x) = f(x), then \int_{a}^{b} x f(x) dx is equal to
(A) \frac{a+b}{2} \int_{a}^{b} f(b-x) dx

(B) \frac{a+b}{2} \int_{a}^{b} f(b+x) dx

(C) \frac{b-a}{2} \int_{a}^{b} f(x) dx

(D) \frac{b+a}{2} \int_{a}^{b} f(x) dx
Sol.
Let I = \int_{a}^{b} x f(x) dx …………….(i)
= \int_{a}^{b} (a + b – x) f (a + b – x) dx
[∵ \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b – x) dx]
= \int_{a}^{b} (a + b – x) f(x) dx
⇒ I = (a + b) \int_{a}^{b} f(x) dx – I [Using Eq.(i)]
⇒ I = (a + b) \int_{a}^{b} f(x) dx
⇒ 2I = (a + b) \int_{a}^{b} f(x) dx
= \left(\frac{a+b}{2}\right) \int_{a}^{b} f(x) d x
Hence, the correct answer is (D).

Question 44.
The value of \int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) dx is
(A) 1
(B) 0
(C) – 1
(D) \frac{\pi}{4}
Solution.

PSEB 12th Class Maths Solutions Chapter 7 Integrals Miscellaneous Exercise 36

Adding Eqs. (i) and (ii), we get
2I = \int_{0}^{1} (tan-1 x + tan-1 (1 – x) – tan-1 (1 – x) – tan-1 x dx
⇒ 2I = 0
⇒ I = 0
Hence, the correct answer is (B).

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