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

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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 e^x – 1 = t²
⇒ 2 e^x 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) ∫ e^x (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 = ∫ e^x (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) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x 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 = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x 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
= \(\left[\frac{x^4}{4}-2 x^3+\frac{9 x^2}{2}+2 x\right]_0^1\)
= \(\left[\frac{1}{4}-2+\frac{9}{2}+2\right]=\frac{19}{4}\)