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

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

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

Evaluate the following:
Question 1.
(i) ∫(1 + x) ex dx
Solution:
∫(1 + x) ex dx
[Choolse 1 + x as first and ex as second function
= (1 + x) ex – ∫1 . ex dx
= ( 1 + x) ex – ex + C = xex + C

(ii) ∫x3 ex dx
Solution:
∫x3 ex dx = x3 ex – ∫3x2 ex dx
= x3 ex – 3{x2 ex – ∫2x ex dx}
= x3 ex – 3x2 ex + 6 ∫x ex dx
= x3 ex – 3x2 ex +6 {x . ex – ∫1 . ex dx}
= x3 ex – 3x2 ex + 6x ex – 6ex + C

(iii) ∫x2 eax dx
Solution:

(iv) ∫(3x + 2)2 e2x dx
Solution:

Question 2.
(i) ∫x sin x dx
Solution:
∫x sin x dx
[x = first function
sin x = 2nd function]
= x (-cosx) – ∫$$\frac{d}{d x}$$(x) . (-cos x) dx
= -x cos x + ∫cos x dx
= -x cos x + sin x + C

(ii) ∫x2 cos x dx
Solution:
∫x2 cos x dx
[x2 = 1st
cos x = 2nd]
= x2 . sin x – ∫$$\frac{d}{d x}$$(x2) sin x dx
= x2 sin x – ∫2x . sin x dx
[x = 1st
sin x = 2nd]
= x2 sin x – 2 {x . (-cos x) – ∫1 . (-cos x) dx}
= x2 sin x + 2x cos x – 2∫cos x dx
= x2 sin x + 2x cos x – 2 sin x + C

(iii) ∫x2 sin ax dx
Solution:

(iv) ∫x cos2 x dx
Solution:

(v) ∫x sin3 x dx
Solution:

(vi) ∫2x sin 2x cos x dx
Solution:

(vii) ∫2x cos 3x cos 2x dx
Solution:

(viii)∫2x3 cos x2 dx
Solution:
∫2x3 cos x2 dx
[Put x2 =t
Then 2x dx = dt]
= ∫x2 . cos x2 . 2x dx
= ∫t . cos t dt
= t . sin t – ∫1 . sin t dt
= t sin t + cos t + C
= x2 sin x2 + cos x2 + C

(ix) ∫x cosec2 x dx
Solution:
∫x cosec2 x dx
[x = 1st
cosec2 x = 2nd]
= x ∫cosec2 x dx – ∫[$$\frac{d}{d x}$$(x) × ∫cosec2 x dx] dx
= -x cot x + ∫cot x dx
= -x cot x + ln |sin x| + C

(x) ∫x tan2 x dx
Solution:
∫x tan2 x dx = ∫x (sec2 x – 1) dx
= ∫x sec2 x dx – ∫x dx
= x tan x – ∫1 . tan x dx – $$\frac{1}{2}$$x2
[x = 1st
sec2 x = 2nd]
= x tan x + ln |cos x| – $$\frac{x^2}{2}$$ + C

Question 3.
(i) ∫x ln (1 + x) dx
Solution:

(ii) ∫x7 ln x dx
Solution:

(iii) ∫(ln x)3 dx
Solution:

(iv) ∫ln(x2 + 1) dx
Solution:
∫ln (x2 + 1) dx
= ∫ln (x2 + 1) . 1 dx
[Put ln (x2 + 1 ) as first function and 1 as the second function.]

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

(vi) ∫ln (x2 + x + 2) dx
Solution:

(vii) ∫ln (x + $$\sqrt{x^2+a^2}$$) dx
Solution:

(viii) ∫ln (x + $$\sqrt{x^2-a^2}$$) dx
Solution:

Question 4.
(i) ∫sin-1 x dx
Solution:

(ii) ∫x sin-1 dx
Solution:

(iii) ∫cos-1 x dx
Solution:

(iv) ∫x tan-1 dx
Solution:

(v) ∫x2 tan-1 x dx
Solution:

(vi) ∫sec-1 x dx
Solution:

(vii) ∫x cosec-1 x dx
Solution:

Question 5.
(i) ∫e3x cos 2x dx
Solution:

(ii) ∫ex sin x dx
Solution:

(iii) ∫ex cos2 x dx
Solution:

(iv) ∫x $$e^{x^2}$$ sin x2 dx
Solution:

(v) ∫eax sin (bx + c) dx
Solution:
Let I = ∫eax sin (bx + c) dx
Integrating by parts we get

(vi) ∫(2x2 + 1)$$e^{x^2}$$ dx
Solution:
I = ∫(2x2 + 1)$$e^{x^2}$$ dx
= ∫2x2 $$e^{x^2}$$ dx + ∫$$e^{x^2}$$ . 1 dx
= ∫2x2 $$e^{x^2}$$ dx + x2 $$e^{x^2}$$ ∫2x$$e^{x^2}$$ .x dx
= x$$e^{x^2}$$ + C

Question 6.
(i) ∫$$\sqrt{9-x^2}$$ dx
Solution:

(ii) ∫$$\sqrt{5-4 x^2}$$ dx
Solution:

(iii) ∫$$\sqrt{1-x^2-2 x}$$ dx
Solution:

(iv) ∫ez $$\sqrt{4-e^{2 z}}$$ dz
Solution:

(v) ∫cos θ $$\sqrt{5-\sin ^2 \theta}$$ dθ
Solution:

Question 7.
(i) ∫$$\sqrt{x^2+4}$$ dx
Solution:

(ii) ∫$$\sqrt{7 x^2+2}$$ dx
Solution:

(iii) ∫$$\sqrt{4 x^2+12 x+13}$$ dx (2x + 3 = z)
Solution:

(iv) ∫e2z $$\sqrt{e^{4 z}+6}$$ dz
Solution:

(v) ∫sec2 θ $$\sqrt{\sec ^2 \theta+3}$$ dθ
Solution:

(vi) ∫(2x2 +1) $$e^{x^2}$$ dx
Solution:
Same as No. 5 (vi).

Question 8.
(i) ∫$$\sqrt{x^2-8}$$ dx
Solution:

(ii) ∫$$\sqrt{3 x^2-2}$$ dx
Solution:

(iii) ∫$$\sqrt{x^2-4 x+2}$$ dx (x – 2 = z)
Solution:

(iv) ∫az $$\sqrt{a^{2 z}-4}$$ dz
Solution:

(v) ∫sec θ tan θ $$\sqrt{\tan ^2 \theta-3}$$ dθ
Solution:

Question 9.
(i)∫ex (tan x + ln sec x) dx
Solution:
∫ex (tan x + ln sec x) dx
= ∫ex tan x dx + ∫ex ln sec x dx
(Integrating by parts)
= ∫ex tan x dx + ex ln sec x – ∫ex tan x dx
= ex ln(sec x) + C

(ii) ∫ex (cot x + ln sin x) dx
Solution:
∫ex (cot x + ln sin x) dx
[Integrating by parts taking ex as first function and cot x as second function.]
= ∫ex ln sin x – ∫ex ln sin x dx + ∫ex ln sin x dx + C
= ex ln sin x + C

(iii) ∫$$\frac{e^x}{x}$$ (1 + x ln x) dx
Solution:
∫$$\frac{e^x}{x}$$ (1 + x ln x) dx
= ∫$$\frac{e^x}{x}$$ dx + ∫$$\frac{e^x}{x}$$ ex ln x dx + C
= ex ln x + C

(iv) ∫$$\frac{x e^x}{(1+x)^2}$$ dx
Solution:

Question 10.
(i) ∫$$\left[\frac{1}{\ln x}-\frac{1}{(\ln x)^2}\right]$$ dx
Solution:

(ii) ∫sin (ln x) dx
Solution:
Let I = ∫sin (ln x) dx
[Integrating by parts taking sin (ln x) as first and 1 as second function.]

(iii) ∫sin x ln (cosec x – cot x) dx
Solution:
∫sin x ln (cosec x – cot x) dx
[Integrating by parts taking In (cosec x cot x) as first function and sin x as second function.]
= ln (cosec x – cot x) . – cos x – ∫$$\frac{1}{{cosec} x-\cot x}$$× – cosec x . cot x + cosec2 x × – cos x dx
= -cos x . ln (cosec x – cot x) + ∫$$\frac{{cosec} x({cosec} x-\cot x)}{{cosec} x-\cot x}$$ . cos x dx
= -cos x . ln (cosec x – cot x) + ∫cot x dx
= -cos x . ln (cosec x – cot x) + ln sin x + C