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) e^{x} dx

Solution:

∫(1 + x) e^{x} dx

[Choolse 1 + x as first and ex as second function

= (1 + x) e^{x} – ∫1 . e^{x} dx

= ( 1 + x) e^{x} – e^{x} + C = xe^{x} + C

(ii) ∫x^{3} e^{x} dx

Solution:

∫x^{3} e^{x} dx = x^{3} e^{x} – ∫3x^{2} e^{x} dx

= x^{3} e^{x} – 3{x^{2} e^{x} – ∫2x e^{x} dx}

= x^{3} e^{x} – 3x^{2} e^{x} + 6 ∫x e^{x} dx

= x^{3} e^{x} – 3x^{2} e^{x} +6 {x . e^{x} – ∫1 . e^{x} dx}

= x^{3} e^{x} – 3x^{2} e^{x} + 6x e^{x} – 6e^{x} + C

(iii) ∫x^{2} e^{ax} dx

Solution:

(iv) ∫(3x + 2)^{2} e^{2x} 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) ∫x^{2} cos x dx

Solution:

∫x^{2} cos x dx

[x^{2} = 1st

cos x = 2nd]

= x^{2} . sin x – ∫\(\frac{d}{d x}\)(x^{2}) sin x dx

= x^{2} sin x – ∫2x . sin x dx

[x = 1st

sin x = 2nd]

= x^{2} sin x – 2 {x . (-cos x) – ∫1 . (-cos x) dx}

= x^{2} sin x + 2x cos x – 2∫cos x dx

= x^{2} sin x + 2x cos x – 2 sin x + C

(iii) ∫x^{2} sin ax dx

Solution:

(iv) ∫x cos^{2} x dx

Solution:

(v) ∫x sin^{3} x dx

Solution:

(vi) ∫2x sin 2x cos x dx

Solution:

(vii) ∫2x cos 3x cos 2x dx

Solution:

(viii)∫2x^{3} cos x^{2} dx

Solution:

∫2x^{3} cos x^{2} dx

[Put x^{2} =t

Then 2x dx = dt]

= ∫x^{2} . cos x^{2} . 2x dx

= ∫t . cos t dt

= t . sin t – ∫1 . sin t dt

= t sin t + cos t + C

= x^{2} sin x^{2} + cos x^{2} + C

(ix) ∫x cosec^{2} x dx

Solution:

∫x cosec^{2} x dx

[x = 1st

cosec^{2} x = 2nd]

= x ∫cosec^{2} x dx – ∫[\(\frac{d}{d x}\)(x) × ∫cosec^{2} x dx] dx

= -x cot x + ∫cot x dx

= -x cot x + ln |sin x| + C

(x) ∫x tan^{2} x dx

Solution:

∫x tan^{2} x dx = ∫x (sec^{2} x – 1) dx

= ∫x sec^{2} x dx – ∫x dx

= x tan x – ∫1 . tan x dx – \(\frac{1}{2}\)x^{2}

[x = 1st

sec^{2} x = 2nd]

= x tan x + ln |cos x| – \(\frac{x^2}{2}\) + C

Question 3.

(i) ∫x ln (1 + x) dx

Solution:

(ii) ∫x^{7} ln x dx

Solution:

(iii) ∫(ln x)^{3} dx

Solution:

(iv) ∫ln(x^{2} + 1) dx

Solution:

∫ln (x^{2} + 1) dx

= ∫ln (x^{2} + 1) . 1 dx

[Put ln (x^{2} + 1 ) as first function and 1 as the second function.]

(v) ∫\(\frac{\ln x}{x^5}\) dx

Solution:

(vi) ∫ln (x^{2} + 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) ∫x^{2} tan^{-1} x dx

Solution:

(vi) ∫sec^{-1} x dx

Solution:

(vii) ∫x cosec^{-1} x dx

Solution:

Question 5.

(i) ∫e^{3x} cos 2x dx

Solution:

(ii) ∫e^{x} sin x dx

Solution:

(iii) ∫e^{x} cos^{2} x dx

Solution:

(iv) ∫x \(e^{x^2}\) sin x^{2} dx

Solution:

(v) ∫e^{ax} sin (bx + c) dx

Solution:

Let I = ∫e^{ax} sin (bx + c) dx

Integrating by parts we get

(vi) ∫(2x^{2} + 1)\(e^{x^2}\) dx

Solution:

I = ∫(2x^{2} + 1)\(e^{x^2}\) dx

= ∫2x^{2} \(e^{x^2}\) dx + ∫\(e^{x^2}\) . 1 dx

= ∫2x^{2} \(e^{x^2}\) dx + x^{2} \(e^{x^2}\) ∫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) ∫e^{z} \(\sqrt{4-e^{2 z}}\) 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) ∫e^{2z} \(\sqrt{e^{4 z}+6}\) dz

Solution:

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

Solution:

(vi) ∫(2x^{2} +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) ∫a^{z} \(\sqrt{a^{2 z}-4}\) dz

Solution:

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

Solution:

Question 9.

(i)∫e^{x} (tan x + ln sec x) dx

Solution:

∫e^{x} (tan x + ln sec x) dx

= ∫e^{x} tan x dx + ∫e^{x} ln sec x dx

(Integrating by parts)

= ∫e^{x} tan x dx + e^{x} ln sec x – ∫e^{x} tan x dx

= e^{x} ln(sec x) + C

(ii) ∫e^{x} (cot x + ln sin x) dx

Solution:

∫e^{x} (cot x + ln sin x) dx

[Integrating by parts taking e^{x} as first function and cot x as second function.]

= ∫e^{x} ln sin x – ∫e^{x} ln sin x dx + ∫e^{x} ln sin x dx + C

= e^{x} 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 + ∫\(\frac{e^x}{x}\) e^{x} ln x dx + C

= e^{x} 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 + cosec^{2} 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