Continuous practice using Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.19 can lead to a stronger grasp of mathematical concepts.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Evaluate the following (1 to 9) integrals :

Question 1.
(i) ∫ $$\frac{x^5}{\sqrt{1+x^3}}$$ dx
(ii) ∫ $$\frac{d x}{\sqrt{2 e^x-1}}$$
Solution:
(i) Let I = ∫ $$\frac{x^5}{\sqrt{1+x^3}}$$ dx
put $$\sqrt{1+x^3}$$ = t
⇒ 1 + x3 = t2
⇒ x3 = t2 – 1
⇒ 3x2 dx = 2t dt
∴ I = ∫ $$\frac{\left(t^2-1\right) 2 t d t}{3 t}$$
= $$\frac{2}{3}$$ ∫ (t2 – 1) dt
= $$\frac{2}{3}\left[\frac{t^3}{3}-t\right]$$
= $$\frac{2}{9}$$ (1 + x3)3/2 – $$\frac{2}{3} \sqrt{1+x^3}$$ + C

(ii) Let I = ∫ $$\frac{d x}{\sqrt{2 e^x-1}}$$
put $$\sqrt{2 e^x-1}$$ = t
⇒ 2 ex – 1 = t2
⇒ 2 ex dx = 2t dt
∴ I = ∫ $$\frac{t d t}{e^x \cdot t}$$
= ∫ $$\frac{\frac{d t}{t^2+1}}{\frac{2}{2}}$$
= 2 ∫ $$\frac{d t}{t^2+1^2}$$
= 2 tan-1 t + C
= 2 tan-1 $$\left(\sqrt{2 e^x-1}\right)$$ + C

Question 2.
(i) ∫ $$\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}$$
(ii) ∫ $$\frac{d x}{\sec x+\ {cosec} x}$$
Solution:
(i) ∫ $$\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}$$

(ii) ∫ $$\frac{d x}{\sec x+\ {cosec} x}$$

Question 3.
(i) ∫ $$\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}$$ dx
(ii) ∫ $$\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}$$ dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ $$\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}$$ dx

(ii) Let I = ∫ $$\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}$$ dx

Question 4.
(i) ∫ $$\frac{d x}{2 \sin x+3 \sec x}$$
(ii) ∫ $$\frac{d x}{\sin ^3 x+\cos ^3 x}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{2 \sin x+3 \sec x}$$

(ii) Let I = ∫ $$\frac{d x}{\sin ^3 x+\cos ^3 x}$$

Question 5.
(i) ∫ $$\frac{x^2-3}{x^3-2 x^2-x+2}$$ dx
(ii) ∫ x tan-1 (2x + 3) dx
Solution:
(i) Let I = ∫ $$\frac{x^2-3}{x^3-2 x^2-x+2}$$ dx
= ∫ $$\frac{\left(x^2-3\right) d x}{(x-1)\left(x^2-x-2\right)}$$
= ∫ $$\frac{\left(x^2-3\right) d x}{(x-1)(x+1)(x-2)}$$
Let $$\frac{x^2-3}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}$$ …………..(1)
Multiply both sides ofeqn. (1) by (x- 1) (x + 1) (x – 2); we get
x2 – 3 = A (x + 1) (x – 2) + B (x – 1) (x – 2) + C (x – 1) (x + 1) …………..(2)
putting x = – 1, 1, 2 successively in eqn. (2) ; we have
– 2 = B (- 2) (- 3)
⇒ B = – $$\frac{1}{3}$$
– 2 = A (2) (- 1)
⇒ A = 1
and1 = 3C
⇒ C = $$\frac{1}{3}$$
∴ from (1) ; we get

(ii) Let I = ∫ x tan-1 (2x + 3) dx

Question 6.
(i) ∫ $$\frac{\tan ^{-1} x}{x^2}$$ dx
(ii) ∫ $$\frac{\log |x|}{(x+1)^3}$$ dx
Solution:
(i) Let I = ∫ tan-1 x . $$\frac{1}{x^2}$$ dx

(ii) Let I = ∫ $$\frac{\log |x|}{(x+1)^3}$$ dx

Multiplying eqn.(2) by x (x + 1)2;
we have 1 = A (x + 1)2 + B x (x + 1) + Cx ………….(3)
puuing x = 0, – 1 successively in eqn.(3);
we have 1 = A and 1 = – C
⇒ C = – 1
Coeff. of x2 ;
0 = A + B
⇒ B = – 1
∴ from eqn. (2); we get

Question 7.
(i) ∫ ex (log x + $$\frac{1}{x^2}$$) dx
(ii) ∫ cos 2x log $$\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)$$ dx
Solution:
(i) Let I = ∫ ex (log x + $$\frac{1}{x^2}$$) dx

(ii) Let I = ∫ cos 2x log $$\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)$$ dx [by parts]

Question 8.
(i) ∫ $$\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}$$ ex dx
(ii) ∫ $$\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}$$ ex dx

(ii) Let I = ∫ $$\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}$$ dx
Divide numerator and denominator by x2 ; we have

Question 9.
$$\int_0^\pi$$ |cos x – sin x| dx
Solution:
Let I = $$\int_0^\pi$$ |cos x – sin x| dx
When 0 ≤ x ≤ $$\frac{\pi}{4}$$ ;
cos x ≥ sin x
⇒ cos x – sin x ≥ 0
∴ |cos x – sin x| = cos x – sin x
When $$\frac{\pi}{4}$$ ≤ x ≤ 7 ;
sin x ≥ cos x
⇒ cos x – sin x ≤ 0
∴ |cos x – sin x| = – (cos x – sin x)
∴ I = $$\int_0^{\pi / 4}$$ |cos x – sin x| dx + $$\int_{\pi / 4}^\pi$$ |cos x – sin x| dx
= $$\int_0^{\pi / 4}$$ (cos x – sin x) dx + $$\int_{\pi / 4}^\pi$$ – (cos x – sin x) dx
= [sin x + cos x$$]_0^{\pi / 4}$$ – [sin x + cos x$$]_{\pi / 4}^\pi$$
= $$\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]-\left[0-1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]$$
= (√2 – 1) – (- 1 – √2)
= 2√2

Question 10.
Prove that $$\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x=\frac{\pi(\pi-\alpha)}{\sin \alpha}$$.
Solution:
Let I = $$\int_0^\pi \frac{x}{1-\cos \alpha \sin x}$$ dx ……………(1)

Question 11.
Prove that $$\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}$$.
Solution:
Let I = $$\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}$$

Question 12.
Evaluate $$\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}$$.
Solution:
Let I = $$\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}$$ ……………(1)

Question 13.
If p (x) is a polynomial of least degree that has maximum value equal to 6 at x = 1, and a minimum value equal to 2 at x = 3, then show that $$\int_0^1$$ p(x) dx = $$\frac{19}{4}$$.
Solution:
Since p (x) be a polynomial of least degree
and it has maximum and minimum values.
∴ p’ (x) must be atleast polynomial of degree 2.
Thus p (x) must be a polynomial of degree 3.
Let p(x) = ax3 + bx2 + cx + d ……………….(1)
∴ p’(x) = 3ax2 + 2bx + c
Since p (x) has maximum values equal to 6 at x = 1
and a minimum value equal to 2 at x = 3.
∴ p'(1) = 0 – p'(3)
⇒ 3a + 2b + c = 0 …………..(2)
and 27a + 6b + c = 0 ……………(3)
also p(1) 6
⇒ a + b + c + d= 6 …………(4)
and p (3) = 2
⇒ 27a + 9b + 3c + d = 2 …………….(5)
eqn. (5) – eqn. (4) gives:
26a + 8b + 2c = – 4
⇒ 13a + 4b + c = – 2 ……………..(6)
eqn. (3) – eqn. (2) gives;
24a + 4b = 0
⇒ 6a + b = 0 …………….(7)
eqn. (3) – eqn. (6) gives
14a + 2b = 2
⇒ 7a + b = 1 ……………(8)
eqn. (8) – eqn. (7) gives;
a = 1
∴ from (7);
b = – 6
∴ from eqn.(6);
13 – 24 + c = – 2
⇒ c = 9
from eqn (4) ;
1 – 6 + 9 + d = 6
⇒ d = 2
∴ from (1) ;
p(x) = x3 – 6x2 + 9x + 2
∴ $$\int_0^1$$ p(x) dx = $$\int_0^1$$ (x3 – 6x2 + 9x + 2) dx
= $$\left[\frac{x^4}{4}-2 x^3+\frac{9 x^2}{2}+2 x\right]_0^1$$
= $$\left[\frac{1}{4}-2+\frac{9}{2}+2\right]=\frac{19}{4}$$