# CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(b)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(b)

Integrate the following:
(In some cases suggestions have been given for substitution.)
(i) ∫sin 3x dx
Solution:

(ii) ∫cos ax dx
Solution:

(iii) ∫cos (2 – 7x) dx
Solution:

(iv) ∫sin $$\frac{x}{2}$$ dx
Solution:

(v) ∫sec2 4x dx
Solution:

(vi) ∫cosec2 $$\frac{x}{3}$$ dx
Solution:

(vii) ∫sec (x + 2) tan (x + 2) dx
Solution:
∫sec (x + 2) tan (x + 2) dx
[Put x + 2 = θ
Then dx = dθ]
= ∫sec θ . tan θ dθ
= sec θ + C = sec (x + 2) + C

(viii) ∫cosec (x + $$\frac{\pi}{4}$$) cot (x + $$\frac{\pi}{4}$$) dx (x + $$\frac{\pi}{4}$$ = z)
Solution:

(ix) ∫x2 cos x3 dx (x3 = z)
Solution:

(x) ∫ex . sec ex tan ex dx (ex = z)
Solution:
∫ex . sec ex . tan ex dx
[Put ex = z
Then ex dx = dz]
= ∫ sec z . tan z dz
= sec z + C = sec ex + C

(xi) ∫$$\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}$$ dx
Solution:

Question 2.
(i) ∫sin x cos x dx (sin x = v)
Solution:
∫sin x cos x dx
[Put sin x = v
Then cos x dx = dv]
= ∫v dv = $$\frac{1}{2}$$v2 + C
= $$\frac{1}{2}$$sin2 x + C

(ii) ∫tan3 x sec2 x dx (tan x = t)
Solution:
∫tan3 x sec2 x dx
[Put tan x = t
Then sec2 x dx = dt]
= ∫t3 dt = $$\frac{1}{4}$$t4 + C
= $$\frac{1}{4}$$tan4 x + C

(iii) ∫$$\frac{{cosec}^2 x}{1+\cot x}$$ dx
Solution:
∫$$\frac{{cosec}^2 x}{1+\cot x}$$ dx
[Put 1 + cot x = t
Then -cosec2 x dx = dt
or, cosec2 x dx = – dt]
= ∫ = -∫ = -ln |t| + C
= -ln |1 + cot x| + C

(iv) ∫$$\frac{\sin x}{\cos ^3 x}$$ dx
Solution:

(v) ∫$$\frac{\cos x}{\sin ^5 x}$$ dx
Solution:

(vi) ∫$$\frac{{cosec}^2 x(\ln x)}{x}$$ dx (In x = z)
Solution:
∫$$\frac{{cosec}^2 x(\ln x)}{x}$$ dx
[Put ln x = z
Then $$\frac{d x}{x}$$ = dz]
= ∫cosec2 z dz
= -cot z + C = -cot (ln x) + C

(vii) ∫$$\sqrt{1-\sin x}$$ cos x dx
Solution:

Question 3.
(i) ∫x$$\sqrt{x^2+3}$$ dx (x2 + 3 = v)
Solution:

(ii) ∫$$\frac{7 d x}{2-3 x}$$ dx
Solution:

(iii) ∫$$\frac{x}{\sqrt{x^2-a^2}}$$ dx
Solution:

(iv) ∫$$\frac{x^2+1}{\left(x^3+3 x+7\right)^3}$$ dx
Solution:

(v) ∫(x4 – 3x2 + 1)4 (2x3 – 3x) dx
Solution:

Question 4.
(i) ∫e3x dx
Solution:
∫e3x dx
[Put 3x = t
Then 3 dx = dt
or, dx = $$\frac{1}{3}$$dt]
= ∫et . $$\frac{1}{3}$$dt = $$\frac{1}{3}$$et + C
= $$\frac{1}{3}$$e3x + C

(ii) ∫e2x+7 dx
Solution:
∫e2x+7 dx
[Put 2x + 7 = t
Then 2 dx = dt
or, dx = $$\frac{1}{2}$$dt]
= ∫et . $$\frac{1}{2}$$dt = $$\frac{1}{2}$$et + C
= $$\frac{1}{2}$$e2x+7 + C

(iii) ∫ex/3 dx
Solution:
[Put $$\frac{x}{3}$$ = t
Then dx = 3 dt
= ∫et . 3dt = 3et + C = 3ex/3 + C

(iv) ∫$$e^{x^3}$$ x2 dx
Solution:

(v) ∫a2x dx
Solution:

(vi) ∫2$$a^{x^2}$$ x dx
Solution:

(vii) ∫e2tanx sec2 x dx
Solution:

(viii) ∫$$\frac{e^x}{\left(e^x-2\right)^2}$$ dx
Solution:

(ix) ∫$$e^{\cos ^2 x}$$ sin 2x dx
Solution:
∫$$e^{\cos ^2 x}$$ sin 2x dx
[Put cos2 x = t
Then 2cos x (-sin x)dx – dt
or, -sin 2x dx = dt
or, sin 2x dx = -dt]
= ∫et (-dt) = et + C = –$$e^{\cos ^2 x}$$ + C

Question 5.
(i)∫$$\frac{\sin ^{-1} x}{\sqrt{1-x^2}}$$ dx
Solution:

(ii)∫$$\frac{d x}{\sqrt{1-(x-1)^2}}$$
Solution:

(iii)∫$$\frac{\left(\sec ^{-1} x\right)^2 d x}{x \sqrt{x^2-1}}$$
Solution:

(iv)∫$$\frac{d x}{x\left[1+(\ln x)^2\right]}$$
Solution:

(v)∫$$\frac{d x}{x^2+2 x+2}$$ (Integrate $$\frac{d x}{(x+1)^2+1}$$)
Solution:

Question 6.
(i) ∫tan 3x dx
Solution:

(ii) ∫cos $$\frac{x}{3}$$ dx
Solution:

(iii) ∫sec (2x + 1) dx
Solution:
∫sec (2x + 1) dx
[Put 2x + 1 = t
Then 2 dx = dt
or, dx = $$\frac{1}{2}$$dt]
= ∫sec t . $$\frac{1}{2}$$dt = $$\frac{1}{2}$$ ∫sec t dt
= $$\frac{1}{2}$$ ln |sec t + tan t| + C
= $$\frac{1}{2}$$ ln |sec (2x + 1) + tan (2x + 1)| + C

(iv) ∫cosec 7x dx
Solution:
∫cosec 7x dx
[Put 7x = t
Then 7 dx = dt
or, dx = $$\frac{1}{7}$$dt]
= ∫cosec t . $$\frac{1}{7}$$dt
= $$\frac{1}{7}$$ ln |cosec t + cot t| + C
= $$\frac{1}{7}$$ ln |cosec 7x + cot x| + C

(v) ∫2x cot (x2 + 3) dx    (x2 + 3 = z)
Solution:
∫2x cot (x2 + 3) dx
[Put x2 + 3 = z
Then 2x dx = dz]
= ∫cot dz
= In |sin z| + C
= In |sin (x2 + 3)| + C

(vi) ∫ex tan ex dx
Solution:
∫ex tan ex dx
[Put ex = t
Then ex dx = dt]
= ∫tan t dt
= In |sec t| + C
= In |sec ex| + C

(vii) ∫(sec 2x – 3)2 dx
Solution:
∫(sec 2x – 3)2 dx
∫sec2 2x dx – 6∫sec 2x dx + 9∫dx
= $$\frac{1}{2}$$ tan 2x – 3 ln |sec 2x + tan 2x| + 9x + C

Question 7.
(i) ∫$$\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ dx
Solution:
∫$$\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ dx
[Put (ex – e-x) = t
Then (ex + e-x)dx = dt]
= ∫$$\frac{d t}{t}$$ = ln |t| + C = ln |(ex – e-x| + C

(ii) ∫3x e2x dx
Solution:

(iii) ∫$$\frac{(x+1) \ln \left(x^2+2 x+2\right)}{x^2+2 x+2}$$ dx
Solution:

Question 8.
(i) ∫$$\frac{\sin x}{\sin (x+\alpha)}$$ dx
Solution:

= ∫(cos α – sin α . cot t) dt
= cos α ∫dt – sin α ∫cot dt
= cos α . t – sin α . ln |sin t| + C
= (x + α) cos α – sin α  ln |sin (x + α)| + C
= x cos α – sin α ln |sin (x + α)| + C

(ii) ∫$$\frac{\sin x}{\cos (x-\alpha)}$$ dx
Solution:

(iii) ∫$$\frac{\tan x+\tan \alpha}{\tan x-\tan \alpha}$$ dx
Solution: