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## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.16

Very Short answer type questions (1 to 5) :

Evaluate the following (1 to 19) integrals :

Question 1.
(i) $$\int_2^3$$ x2 dx (NCERT)
(ii) $$\int_{-1}^1$$ (x + 1) dx (NCERT)
(iii) $$\int_2^3 \frac{1}{x}$$ dx
Solution:
(i) $$\int_2^3$$ x2 dx
= $$\frac{1}{3}$$ (33 – 23)
= $$\frac{1}{3}$$ (27 – 8)
= $$\frac{19}{3}$$

(ii) $$\int_{-1}^1$$ (x + 1) dx
= $$\frac{1}{2}$$ [(1 + 1)2 – (- 1 + 1)2]
= $$\frac{1}{2}$$ [4 – 0]
= 2

(iii) $$\int_2^3 \frac{1}{x}$$ dx = log |x|$$]_2^3$$
= log 3 – log 2
= log $$\frac{3}{2}$$

Question 2.
(i) $$\int_{-4}^{-1} \frac{1}{x}$$ dx
(ii) $$\int_0^1 \frac{1}{2 x-3}$$ dx
(iii) $$\int_0^1 \sqrt{5 x+4}$$ dx
Solution:
(i) $$\int_{-4}^{-1} \frac{1}{x}$$ dx
= log $$|x|]_{-4}^{-1}$$
= log |- 1| – log |- 4|
= 0 – log 4
= – log 4

(ii) $$\left.\int_0^1 \frac{1}{2 x-3} d x=\frac{\log |2 x-3|}{2}\right]_0^1$$
= $$\frac{1}{2}$$ [log |2 – 3| – log |0 – 3|]
= $$\frac{1}{2}$$ [log |- 1| – log |- 3|]
= $$\frac{1}{2}$$ [0 – log 3]
= – $$\frac{1}{2}$$ log 3

(iii) $$\int_0^1 \sqrt{5 x+4}$$ dx
= $$\left.\frac{(5 x+4)^{\frac{1}{2}+1}}{5\left(\frac{1}{2}+1\right)}\right]_0^1$$
= $$\left.\frac{2}{15}(5 x+4)^{3 / 2}\right]_0^1$$
= $$\frac{2}{15}$$ [93/2 – 43/2]
= $$\frac{2}{15}$$ [27 – 8]
= $$\frac{38}{15}$$

Question 3.
(i) $$\int_2^3$$ 3x dx (NCERT)
(ii) $$\int_0^{\pi / 4}$$ tan x dx (NCERT)
(iii) $$\int_{\pi / 4}^{\pi / 2}$$ cot x dx
Solution:
(i) Let I = $$\int_2^3$$ 3x dx
= $$\left.\frac{3^x}{\log 3}\right]_2^3$$
= $$\frac{1}{\log 3}$$ [33 – 32]
= $$\frac{18}{\log 3}$$

(ii) $$\int_0^{\pi / 4}$$ tan x dx
= $$\int_0^{\pi / 4} \frac{\sin x}{\cos x}$$ dx
= – log |cos x|$$]_0^{\pi / 4}$$
= – log $$\frac{1}{\sqrt{2}}$$
= – log 2– 1/2
= $$\frac{1}{2}$$ log 2

(iii) $$\int_{\pi / 4}^{\pi / 2}$$ cot x dx
= $$\int_{\pi / 4}^{\pi / 2} \frac{\cos x}{\sin x}$$ dx
= log |sin x| $$]_{\pi / 4}^{\pi / 2}$$
= log sin $$\frac{\pi}{2}$$ – log sin $$\frac{\pi}{4}$$
= 0 – log 2– 1/2
= $$\frac{1}{2}$$ log 2.

Question 4.
(i) $$\int_0^{\pi / 4}$$ sin 2x dx (NCERT)
(ii) $$\int_0^{\pi / 2}$$ cos 2x dx (NCERT)
(iii) $$\int_0^\pi$$ (sin2 $$\frac{x}{2}$$ – cos2 $$\frac{x}{2}$$) dx (NCERT)
Solution:
(i) $$\int_0^{\pi / 4}$$ sin 2x dx
= – $$\left.\frac{\cos 2 x}{2}\right]_0^{\pi / 4}$$
= – $$\frac{1}{2}$$ [cos $$\frac{\pi}{2}$$ – cos 0]
= – $$\frac{1}{2}$$ (0 – 1)
= $$\frac{1}{2}$$

(ii) $$\int_0^{\pi / 2}$$ cos 2x dx
= $$\left.\frac{\sin 2 x}{2}\right]_0^{\pi / 2}$$
= $$\frac{1}{2}$$ [sin π – sin 0] = 0

(iii) Let I = $$\int_0^\pi$$ (sin2 $$\frac{x}{2}$$ – cos2 $$\frac{x}{2}$$) dx
= $$\int_0^\pi$$ – cos (2 × $$\frac{x}{2}$$) dx
[∵ cos 2θ = cos2 θ – sin2 θ]
= – $$\int_0^\pi$$ cos x dx
= – [sin x]0 π
= – [sin π – sin 0]
= – (0 – 0) = 0

Question 5.
(i) $$\int_0^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}$$
(ii) $$\int_0^3 \frac{d x}{9+x^2}$$
(iii) $$\int_1^{\sqrt{3}} \frac{d x}{1+x^2}$$
Solution:
(i) $$\left.\int_0^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}=\sin ^{-1} x\right]_0^{1 / \sqrt{2}}$$
= sin-1 $$\left(\frac{1}{\sqrt{2}}\right)$$ – sin-1 0
= $$\frac{\pi}{4}$$ – 0
= $$\frac{\pi}{4}$$

(ii) $$\int_0^3 \frac{d x}{9+x^2}$$

(iii) $$\left.\int_1^{\sqrt{2}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}$$
= tan-1 √3 – tan-1 1
= $$\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$
[∵ tan $$\frac{\pi}{4}$$ = 1
and tan $$\frac{\pi}{3}$$ = √3]

Question 6.
(i) $$\int_2^3 \frac{d x}{x^2-1}$$ (NCERT)
(ii) $$\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x}$$ dx
Solution:
(i) $$\int_2^3 \frac{d x}{x^2-1}$$
= $$\left.\frac{1}{2 \times 1} \log \left|\frac{x-1}{x+1}\right|\right]_2^3$$
= $$\frac{1}{2}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{3}\right)\right]$$
= $$\frac{1}{2} \log \frac{3}{2}$$

(ii) $$\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x}$$

Question 6 (old).
(ii) $$\int_0^\pi \frac{1}{1+\sin x} d x$$
Solution:
Let I = $$\int_0^\pi \frac{1}{1+\sin x} d x$$

= [(tan π – sec π) – (tan 0 – sec 0)]
= [0 – (- 1)] – (0 – 1)
= 1 + 1 = 2.

Question 7.
(i) $$\int_1^2$$ (4x3 – 5x2 + 6x + 9) dx (NCERT)
(ii) $$\int_0^8\left(\sqrt{8 x}-\frac{x^2}{8}\right)$$ dx
Solution:
(i) $$\int_1^2$$ (4x3 – 5x2 + 6x + 9) dx

(ii) $$\int_0^8\left(\sqrt{8 x}-\frac{x^2}{8}\right)$$ dx

Question 8.
(i) $$\int_{\pi / 6}^{\pi / 4}$$ cosec x dx (NCERT)
(ii) $$\int_0^{\pi / 4}$$ (2 sec2 x + x3 + 2) dx (NCERT)
Solution:
(i) $$\int_{\pi / 6}^{\pi / 4}$$ cosec x dx
= log |cosec x – cot x|$$]_{\pi / 6}^{\pi / 4}$$
= 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 |√2 – 1| – log |2 – √3|

(ii) $$\int_0^{\pi / 4}$$ (2 sec2 x + x3 + 2) dx

Question 9.
(i) $$\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}}$$ (NCERT)
(ii) $$\int_0^1 \frac{1-x}{1+x}$$ dx
Solution:
(i) Let I = $$\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}}$$

(ii) Let I = $$\int_0^1 \frac{1-x}{1+x}$$ dx
= – $$\int_0^1 \frac{x-1}{x+1}$$ dx
= – $$\int_0^1 \frac{x+1-2}{x+1}$$ dx
= – $$\int_0^1\left[1-\frac{2}{x+1}\right]$$ dx
= – [x – 2 log |x + 1|$$]_0^1$$
= – [(1 – 2 log |1 + 1|) – (0 – 2 log |0 + 1|)]
= – 1 + 2 log 2 – (0 – 2 log 1)
= – 1 + 2 log 2 – 0
= – 1 + 2 log 2

Question 10.
(i) $$\int_0^{\pi / 2}$$ sin3 x dx
(ii) $$\int_0^{\pi / 4}$$ (tan x + cot x)-1 dx (ISC 2003)
Solution:
(i) I = $$\int_0^{\pi / 2}$$ sin3 x dx
= $$\int_0^{\pi / 2}$$ sin2 x sin x dx
= $$\int_0^{\pi / 2}$$ (1 – cos2 x) sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0
⇒ t = cos 0 = 1
and when x = π/2
⇒ t = cos π/2 = 0
∴ I = $$\int_1^0$$ (1 – t2) (- dt)
= – $$\left[t-\frac{t^3}{3}\right]_1^0$$
= $$\left[\frac{t^3}{3}-t\right]_1^0$$
= (0 – 0) – ($$\frac{1}{3}$$ – 1)
= $$\frac{2}{3}$$

(ii) Let I = $$\int_0^{\pi / 4}$$ (tan x + cot x)-1 dx

Question 11.
(i) $$\int_0^{\pi / 2} \sqrt{1-\cos 2 x}$$ dx
(ii) $$\int_0^\pi \sqrt{1+\sin x}$$ dx
Solution:
(i) $$\int_0^{\pi / 2} \sqrt{1-\cos 2 x}$$ dx

= – √2 (cos $$\frac{\pi}{2}$$ – cos 0)
= – √2 (0 – 1)
= √2

(ii) I = $$\int_0^\pi \sqrt{1+\sin x}$$ dx

Question 12.
(i) $$\int_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}$$ dx
(ii) $$\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}$$ dx
Solution:
(i) Let I = $$\int_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}$$ dx

(ii) Let I = $$\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}$$ dx

Question 13.
(i) $$\int_0^{\pi / 4}$$ sin 2x sin 3x dx
(ii) $$\int_0^{\pi / 2}$$ (a2 cos2 x + b2 sin2 x) dx
Solution:
(i) Let I = $$\int_0^{\pi / 4}$$ sin 2x sin 3x dx
= $$\frac{1}{2}$$ $$\int_0^{\pi / 4}$$ (2 sin 3x sin 2x) dx
= $$\frac{1}{2}$$ $$\int_0^{\pi / 4}$$ (cos x – cos 5x) dx
= $$\frac{1}{2}\left[\sin x-\frac{\sin 5 x}{5}\right]_0^{\pi / 4}$$
= $$\frac{1}{2}\left[\sin \frac{\pi}{4}-\frac{1}{5} \sin \frac{5 \pi}{4}-0+0\right]$$
= $$\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{5 \sqrt{2}}\right)=\frac{3}{5 \sqrt{2}}$$

(ii) Let I = $$\int_0^{\pi / 2}$$ (a2 cos2 x + b2 sin2 x) dx

Question 14.
(i) $$\int_{\pi / 3}^{\pi / 4}$$ (tan x + cot x)2 dx
(ii) $$\int_1^2 \frac{d x}{(x+1)(x+2)}$$
Solution:
(i) Let I = $$\int_{\pi / 3}^{\pi / 4}$$ (tan x + cot x)2 dx

(ii) $$\int_1^2 \frac{d x}{(x+1)(x+2)}$$

Question 15.
(i) $$\int_1^2 \frac{x+3}{x(x+2)}$$ dx
(ii) $$\int_1^3 \frac{d x}{x^2(x+1)}$$ (NCERT)
Solution:
(i) Let I = $$\int_1^2 \frac{x+3}{x(x+2)}$$ dx
= $$\frac{1}{2} \int_1^2 \frac{2 x+6}{x^2+2 x}$$ dx

(ii) Let $$\frac{1}{x^2(x+1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{\mathrm{C}}{x+1}$$ ……………….(1)
Multiply eq. (1) by x2 (x + 1) ; we have
1 = Ax (x + 1) + B ( x + 1) + Cx2 ………………..(2)
putting x = 0 in eqn. (2) ; we have 1 = B
puiting x = – 1 ineqn. (2); we have 1 = C
coeff. of x2 ;
0 = A + C = A – 1
∴ from (1); we have

Question 15 (old).
(i) $$\int_1^2 \frac{d x}{(x+1)\left(x^2-7 x+12\right)}$$
(ii) $$\int_1^2 \frac{5 x^2}{x^2+4 x+3}$$ dx
Solution:
(i) Let $$\frac{1}{(x+1)\left(x^2-7 x+12\right)}=\frac{1}{(x+1)(x-3)(x-4)}$$
= $$\frac{A}{x+1}+\frac{B}{x-3}+\frac{C}{x-4}$$ ………………..(1)
Multiplying both sides of eqn. (1) by (x + 1) (x – 3) (x – 4) ; we have
1 = A (x – 3) (x – 4) + B (x + 1) (x – 4) + C (x + 1) (x – 3) …………….(2)
putting x = – 1, 3, 4 successively in eqn. (2) ; we have
1 = A (- 4) (- 5)
⇒ A = $$\frac{1}{20}$$
1 = B (4) (- 1)
⇒ B = – $$\frac{1}{4}$$
and 1 = C (5) . 1
⇒ C = $$\frac{1}{5}$$
∴ from (1) ; we have

(ii) $$\int_1^2 \frac{5 x^2}{x^2+4 x+3}$$ dx
= $$\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right]$$ dx

Question 16.
(i) $$\int_0^{\pi / 2} \frac{5 \sin x+3 \cos x}{\sin x+\cos x}$$ dx
(ii) $$\int_0^{\pi / 2} \frac{3 \sin x+4 \cos x}{\sin x+\cos x}$$ dx
Solution:
(i) Numerator = l (denominator) + m $$\frac{d}{d x}$$ (deno.)
∴ 5 sin x + 3 cos x = l (sin x + cos x) + m $$\frac{d}{d x}$$ (sin x + cos x)
⇒ 5 sin x + 3 cos x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x ;
5 = l – m;
Coeff. of cos x ;
3 = l + m
On solving these eqn’s; we have
l = 4, m = – 1

(ii) Let I = $$\int_0^{\pi / 2} \frac{3 \sin x+4 \cos x}{\sin x+\cos x}$$ dx
= $$\int_0^{\pi / 2}\left[3+\frac{\cos x}{\sin x+\cos x}\right]$$ dx
= $$3 \int_0^{\pi / 2} d x+\int_0^{\pi / 2} \frac{\cos x d x}{\sin x+\cos x}$$ dx
= 3 $$\frac{\pi}{2}$$ + I1 ………………(*)
Let cos x = l (sin x + cos x) + m (cos x – sin x) ………..(1)
Coeff. of cos x ;
1 = l + m ………………(2)
Coeff. of sin x ;
0 = l – m ………………..(3)
On solving (2) and (3) ; we have
l = $$\frac{1}{2}$$ = m
On dividing eqn. (1) throughout by sin x + cos x ; we have

Question 16 (old).
(ii) $$\int_0^\pi \frac{\sin x}{\sin x+\cos x}$$ dx
Solution:
Numerator = l (deno.) + m $$\frac{d}{d x}$$ (deno.)
sin x = l (sin x + cos x) + m $$\frac{d}{d x}$$ (sin x + cos x)
sin x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x;
1 = l – m;
Coeff. of cos x;
0 = l + m
On solving these eqn’s,
l = $$\frac{1}{2}$$ ; m = – $$\frac{1}{2}$$
∴ $$\int_0^\pi \frac{\sin x d x}{\sin x+\cos x}=\int_0^\pi \frac{l(\sin x+\cos x)}{\sin x+\cos x} d x+m \int_0^\pi \frac{(\cos x-\sin x) d x}{\sin x+\cos x}$$
= l $$\int_0^\pi$$ dx + m log |sin x + cos x|$$]_0^\pi$$
= $$\frac{1}{2}$$ × π + (- $$\frac{1}{2}$$) [log 1 – log 1]
= $$\frac{\pi}{2}$$

Question 17.
(i) $$\int_0^1$$ x ex dx (NCERT)
(ii) $$\int_0^1$$ (x esup>x + sin $$\frac{\pi}{4}$$ x) dx (NCERT)
Solution:
(i) $$\int_0^1$$ x ex dx
= $$\left.x e^x\right]_0^1-\int_0^1 e^x d x$$
= $$\left[x e^x-e^x\right]_0^1$$
= (1 . e1 – e1) – (0 – e0)
= 1

(ii) $$\int_0^1$$ (x ex + sin $$\frac{\pi}{4}$$ x) dx

Question 18.
(i) $$\int_0^1\left(x e^{2 x}+\sin \frac{\pi x}{2}\right)$$ dx
(ii) $$\int_0^{\pi / 2}$$ x2 cos 2x dx
Solution:
(i) Let I = $$\int_0^1\left(x e^{2 x}+\sin \frac{\pi x}{2}\right)$$ dx

(ii) $$\int_0^{\pi / 2}$$ x2 cos 2x dx

Question 19.
(i) $$\int_1^2 e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)$$ dx
(ii) $$\int_{\pi / 2}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right)$$ dx (NCERT)
Solution:
(i) Let I = $$\int_1^2 e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)$$ dx

(ii) Let I = $$\int_{\pi / 2}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right)$$ dx

Question 20.
(i) If $$\int_0^a$$ 3x2 dx = 8, find the value of a.
(ii) If $$\int_1^a$$ (3x2 + 2x + 1) dx = 11, find the value(s) of a.
Solution:
(i) Given $$\int_0^a$$ 3x2 dx = 8
⇒ 3 $$\left[\frac{x^3}{3}\right]_0^a$$ = 8
⇒ a3 – 0 = 8 = 23
⇒ a3 – 23 = 0
⇒ (a – 2) (a2 + 2a + 4) = 0
⇒ a = 2, $$\frac{-2 \pm \sqrt{4-16}}{2}$$
i.e., a = 2, $$\frac{-2 \pm \sqrt{4-16}}{2}$$ ; i.e., a = 2, – 1 ± √3 i
Thus, only real value of a be 2.

(ii) $$\int_1^a$$ (3x2 + 2x + 1) dx = 11
⇒ $$\left.\frac{3 x^3}{3}+\frac{2 x^2}{2}+x\right]_1^a$$ = 11
⇒ (a3 + a2 + a) – (1 + 1 + 1) = 11
⇒ a3 + a2 + a – 14 = 0
⇒ (a – 2) (a2 + 3a + 7) = 0
either a – 2 = 0 or a2 + 3a + 17 = 0
it does not gives real values of a
⇒ a = 2
Hence a = 2.

Question 21.
(i) If $$\int_0^a$$ √x dx = 4a $$\int_0^{\pi / 4}$$ sin 2x dx, find the value of a.
(ii) If $$\int_a^b$$ x3 dx = 0 and $$\int_a^b$$ x2 dx = $$\frac{2}{3}$$, find the values of a and b.
Solution:
(i) Given, $$\int_0^a$$ √x dx

(ii) Given $$\int_a^b$$ x3 dx = 0
⇒ $$\left.\frac{x^4}{4}\right]_a^b$$ = 0
⇒ $$\frac{1}{4}$$ (b4 – a4) = 0
⇒ b4 – a4 = 0
⇒ (b – a) (b + a) (b2 + a2) = 0 ………………(1)
and $$\int_a^b$$ x2 dx = $$\frac{2}{3}$$
⇒ $$\left.\frac{x^3}{3}\right]_a^b=\frac{2}{3}$$
⇒ $$\frac{b^3-a^3}{3}=\frac{2}{3}$$
⇒ (b – a) (b2 + ab + a2) = 2 …………..(2)
When b – a = 0 ⇒ b = a, does not satisfies given integrals.
from (1) ; when b + a = 0 ⇒ b = – a
∴ from (2); we have
– 2a (a2 – a2 + a2) = + 2
⇒ a3 = – 1
⇒ a = – 1 [other two values of a are complex numbers]
and b = – a + 1
and other eqn’s does not gives real values of a and b.
Thus, a = – 1 and b = + 1.