Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e2x dx, x ∈ R.
Solution:
∫e2x dx = $$\frac{e^{2x}}{2}$$ + C

Question 2.
∫sin 7x dx, x ∈ R.
Solution:
∫sin 7x dx = $$\frac{\cos 7x}{7}$$ + C

Question 3.
∫$$\frac{x}{1+x^2}$$ dx, x ∈ R.
Solution:
∫$$\frac{xdx}{1+x^2}$$ dx = $$\frac{1}{2}$$∫$$\frac{2xdx}{1+x^2}$$ = $$\frac{1}{2}$$ log(1+x²) + C

Question 4.
∫2xsin(x²+1) dx, x ∈ R.
Solution:
∫2x.sin(x²+1) dx, x ∈ R.
t = x² + 1 ⇒ dt = 2x dx
∫2x. sin(x²+1) dx = ∫ sin t dt = -cos t + C
= -cos (x²+1) + C

Question 5.
∫$$\frac{(logx)^2}{x}$$ dx on I ⊂ (0, ∞).
Solution:
∫$$\frac{(logx)^2}{x}$$ dx
t = log x ⇒ dt = $$\frac{1}{x}$$ dx
∫$$\frac{(logx)^2}{x}$$ dx = ∫t² dt
= $$\frac{(t^3}{3}$$ + C = $$\frac{(logx)^3}{3}$$ + C

Question 6.
dx on I ⊂ (1, ∞).
Solution:

Question 7.
∫$$\frac{\sin(Tan^{-1}x)}{1+x^2}$$ dx, x ∈ R.
Solution:
∫$$\frac{\sin(tan^{-1}x)}{1+x^2}$$ dx
t = tan-1 x ⇒ dt = $$\frac{dx}{1+x^2}$$
∫$$\frac{\sin(tan^{-1}x)}{1+x^2}$$dx = ∫ sin t
= -cos t + t
= -cos (tan-1 x) + C

Question 8.
∫$$\frac{1}{8+2x^2}$$ dx on R.
Solution:

Question 9.
∫$$\frac{3x^2}{1+x^6}$$dx, on R.
Solution:

Question 10.
∫$$\frac{2}{\sqrt{25+9x^2}}$$dx on R.
Solution:

Question 11.
∫$$\frac{3}{\sqrt{9x^2-1}}$$dx on ($$\frac{1}{3}$$, ∞).
Solution:

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx cos nx = $$\frac{1}{2}$$∫ 2 sin mx. cos nx dx
= $$\frac{1}{2}$$∫sin(m+n)x + sin(m-n)x) dx
= –$$\frac{1}{2}$$($$\frac{\cos (m+n) x}{m+n}+ cos\frac{(m-n)x}{m-n}$$) + C

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫sin mx. sin nx dx = $$\frac{1}{2}$$ ∫2 sin mx. sin nx dx
= $$\frac{1}{2}$$∫cos (m – n)x – cos (m + n)x dx
= $$\frac{1}{2}$$($$\frac{\sin (m-n) x}{m-n}+\frac{\sin (m+n)x}{m+n}$$) + C

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
∫cos mx. cos nx dx = $$\frac{1}{2}$$ ∫2 cos mx.cos nx dx
= $$\frac{1}{2}$$∫cos (m + n)x – cos (m – n)x dx
= $$\frac{1}{2}$$($$\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n)x}{m-n}$$) + C

Question 15.
∫sin x sin 2x. sin 3x dx on R.
Solution:
sin 2x . sin 3x = $$\frac{1}{2}$$(2 sin 3x. sin 2x)
= $$\frac{1}{2}$$(cos x – cos 5x)
sin x sin 2x sin 3x
= $$\frac{1}{2}$$(sin x . cos x – cos 5x . sin x)
= $$\frac{1}{2}$$($$\frac{1}{2}$$ sin 2x – $$\frac{1.2}{2}$$cos 5x . sin x)
= $$\frac{1}{2}$$($$\frac{1}{2}$$ sin 2x – $$\frac{1}{2}$$ – (sin 6x – sin 4x)

Question 16.
∫$$\frac{\sin x}{\sin(a+x)}$$ dx on I ⊂ R\{nπ – a : n ∈ Z}.
Solution:
sin x = sin (a + x – a)
= sin (a + x) . cos a – cos (a + x) sin a
∫$$\frac{\sin x}{\sin(a+x)}$$dx
= cos a ∫ dx – sin a ∫$$\frac{\cos (a+x)}{\sin(a+x)}$$ dx
= x . cos a – sin a . log |sin (a + x)| + c

II. Evaluate the following integrals.

Question 1.
∫(3x – 2)½ dx on ($$\frac{2}{3}$$, ∞)
Solution:
t = 3x – 2 ⇒ dt = 3 dx
∫(3x – 2)½ dx = $$\frac{1}{3}$$ ∫t½ dt = $$\frac{1}{3}$$ $$\frac{t^{3/2}}{3/2}$$ + C
= $$\frac{2}{9}$$(3x – 2)3/2 + C

Question 2.
∫$$\frac{1}{7x+3}$$ dx on I ⊂ R\{-$$\frac{3}{7}$$}.
Solution:
∫$$\frac{1}{7x+3}$$ dx
t = 7x + 3
⇒ dt = 7 dx
= ∫$$\frac{1}{7x+3}$$ dx = $$\frac{1}{7}$$ ∫$$\frac{dt}{t}$$
= $$\frac{1}{7}$$ log |t| + C = $$\frac{1}{7}$$ log|7x + 3| + C

Question 3.
∫$$\frac{log(1+x)}{1+x}$$ dx on (-1, ∞).
Solution:
∫$$\frac{log(1+x)}{1+x}$$ dx
t = 1 + x ⇒ dt = dx

Question 4.
∫(3x² – 4)x dx on R.
Solution:
∫(3x² – 4)x dx
t = 3x² – 4 ⇒ dt = 6x dx
∫(3x² – 4)x dx = $$\frac{1}{6}$$∫t dt = $$\frac{1}{6}$$.$$\frac{t^2}{2}$$ + C
= $$\frac{(3x^2-4)^2}{12}$$ + C

Question 5.
∫$$\frac{dx}{\sqrt{1+5x}}$$ dx on (-$$\frac{1}{5}$$, ∞).
Solution:

Question 6.
∫(1 – 2x³)x² dx on R.
Solution:
∫(1 – 2x³)x² dx
t = 1 – 2x³ ⇒ dt = -6x² dx
∫(1 – 2x³)x² dx = –$$\frac{1}{6}$$∫t dt
= –$$\frac{1}{6}$$ . $$\frac{t^2}{2}$$ + C
= $$\frac{-(1-2x^3)^2}{12}$$ + C

Question 7.
∫$$\frac{\sec^2 x}{(1+tan x)^3}$$ dx on I ⊂ R \ {nπ – $$\frac{\pi}{4}$$ : n ∈ Z}.
Solution:
∫$$\frac{\sec^2 x}{(1+tan x)^3}$$ dx
t = 1 + tan x ⇒ dt = sec² x dx

Question 8.
∫x³ sinx4 dx on R.
Solution:
∫x³ . sinx4 dx
t = x4 ⇒ dt = 4x³ dx
∫x³ . sinx4 dx = $$\frac{1}{4}$$∫sin t. dt
= –$$\frac{1}{4}$$cos t + C
= –$$\frac{1}{4}$$. cos x4 + C

Question 9.
∫$$\frac{\cos x}{(1+sin x)^2}$$ dx on I ⊂ R\{2nπ + $$\frac{3 \pi}{2}$$ : n ∈ Z}.
Solution:

Question 10.
∫$$\sqrt[3]{sin x}$$ cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:

Question 11.
∫ 2x e dx on R.
Solution:
∫ 2x. e dx
t = x² ⇒ dt = 2x dx
∫ 2xe dt = ∫ et dt = et + C
= e + C

Question 12.
∫$$\frac{e^{log x}}{x}$$ dx on (0, ∞).
Solution:
∫$$\frac{e^{log x}}{x}$$ dx
t = log x ⇒ dt = $$\frac{1}{x}$$.dx
∫$$\frac{e^{log x}}{x}$$ dx = ∫et. dt
= et + C
= elog x + C
= x + C

Question 13.
∫$$\frac{x^2}{\sqrt{1-x^6}}$$ dx on I = (-1, 1).
Solution:

Question 14.
∫$$\frac{2x^3}{1+x^8}$$ dx on R.
Solution:
t = x4 ⇒ dt = 4x³ dx
∫$$\frac{2x^3}{1+x^8}$$ = $$\frac{1}{2}$$∫$$\frac{dt}{1+t^2}$$
= $$\frac{1}{2}$$ tan-1 t + C
= $$\frac{1}{2}$$ tan-1(x4) + C

Question 15.
∫$$\frac{x^8}{1+x^18}$$ dx on R.
Solution:
t = x9 ⇒ dt = 9x8 dx

Question 16.
∫$$\frac{e^x(1+x)}{\cos^2(xe^x)}$$ dx on I ⊂ R/{x ∈ R : cos (xex) = 0}.
Solution:
t = x . ex
dt = (x . ex + ex)dx = ex(1 + x)dx
∫$$\frac{e^x(1+x)}{\cos^2(xe^x)}$$ dx = ∫$$\frac{dt}{\cos^2 t}$$ dx
= ∫sin² t dt
= tan t + C
= tan (x. ex) = C

Question 17.
∫$$\frac{cosec^2 x}{(a+b cot x)^5}$$ dx on I ⊂ R\ {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
Let t = a + b cot x
dt = -b cosec² x dx
∫$$\frac{cosec^2 x}{(a+b cot x)^5}$$ dx = –$$\frac{1}{b}$$ ∫$$\frac{bt}{t^5}$$
= –$$\frac{1}{b}$$ ∫t-5 dt
= –$$\frac{1}{b}$$ $$\frac{t^{-4}}{-4}$$ + C
= $$\frac{1}{4b t^{4}}$$ + C
= $$\frac{1}{4b(a+b cotx)^{4}}$$ + C

Question 18.
∫ex sin ex dx on R.
Solution:
t = ex ⇒ dt = ex dx
∫ex . sin ex dx = ∫ sin t dt
= -cos t + C
= -cos (ex) + C

Question 19.
∫$$\frac{\sin(log x)}{x}$$ dx on (-1, ∞).
Solution:
t = log x ⇒ dt = $$\frac{1}{x}$$ dx
∫$$\frac{\sin(log x)}{x}$$dx = ∫sin t dt
= -cos t + C
= -cos (log x) + C

Question 20.
∫$$\frac{1}{x log x}$$ dx on (-1, ∞).
Solution:
t = log x
dt = $$\frac{1}{x}$$ . dx
∫$$\frac{1}{x log x}$$ dx = ∫$$\frac{1}{t}$$dt = log t + C = log(log x + C)

Question 21.
∫$$\frac{(1+log x)^n}{x}$$ dx on (e-1, ∞). n ≠ -1.
Solution:
t = 1 + log x
dt = $$\frac{1}{x}$$ dx

Question 22.
∫$$\frac{\cos(log x)}{x}$$ dx on (0, ∞).
Solution:
t = log x
dt = $$\frac{1}{x}$$ dx
∫$$\frac{\cos(log x)dx}{x}$$ = ∫cos t dt
= sin t + C
= sin (log x) + C

Question 23.
∫$$\frac{\cos \sqrt{x}}{\sqrt{x}}$$ dx on (0, ∞).
Solution:

Question 24.
∫$$\frac{2x+1}{x^2+x+1}$$ dx on R.
Solution:

Question 25.
∫$$\frac{ax^{n-1}}{bx^n+c}$$ dx, where n ∈ N, a, b, c are real nembers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xn ≠ –$$\frac{c}{b}$$}.
Solution:
t = bxn + C
dt = nbxn-1dx

Question 26.
∫$$\frac{1}{x log x[log(log x)]}$$ dx on (1, ∞).
Solution:

Question 27.
∫cot hx dx on R.
Solution:
t = sinh x ⇒ dt = cosh x dx
∫cot hx dx = ∫$$\frac{dt}{t}$$
= log |t| +C
= log |log(log x)| + C

Question 28.
∫$$\frac{1}{\sqrt{1-4x^2}}$$ dx on (-$$\frac{1}{2}$$, $$\frac{1}{2}$$).
Solution:

Question 29.
∫$$\frac{dx}{\sqrt{25+x^2}}$$ dx on R.
Solution:

Question 30.
∫$$\frac{1}{(x+3\sqrt{x+2}}$$ dx on I ⊂ (-2, ∞).
Solution:
x + 2 = t²
dx = 2t dt

Question 31.
∫$$\frac{1}{1+\sin 2x}$$ dx on I ⊂ R\{$$\frac{n \pi}{2}$$ + (-1)n $$\frac{\pi}{4}$$ : n ∈ Z}.
Solution:

Question 32.
∫$$\frac{x^2+1}{x^4+1}$$ dx on R.
Solution:
∫$$\frac{x^2+1}{x^4+1}$$ dx
Dividing Nr and Dr by x²

Question 33.
∫$$\frac{dx}{\cos^2+\sin 2x}$$ on I ⊂ R$${(2n+1)\(\frac{\pi}{2}$$ : n ∈ Z}∪{2nπ + tan-1 $$\frac{1}{2}$$ : n ∈ Z})
Solution:
∫$$\frac{dx}{\cos^2+\sin 2x}$$
Dividing Nr and Dr by cos² x

Question 34.
∫$$\sqrt{1-sin 2x}$$ dx on I ⊂ {2nπ – $$\frac{3 \pi}{4}$$, 2nπ + $$\frac{\pi}{4}$$}, n ∈ Z.
Solution:

Question 35.
∫$$\sqrt{1+cos 2x}$$ dx I ⊂ {2nπ – $$\frac{\pi}{2}$$, 2nπ + $$\frac{\pi}{2}$$}, n ∈ Z.
Solution:
∫$$\sqrt{1+cos 2x}$$ dx = ∫$$\sqrt{2cos^2 x}$$ dx
= √2 ∫ cos x dx
= √2 sin x + C

Question 36.
∫$$\frac{\cos x + \sin x}{\sqrt{1+\sin 2x}}$$ dx on I ⊂ {2nπ – $$\frac{\pi}{4}$$, 2nπ + $$\frac{3\pi}{4}$$}, n ∈ Z.
Solution:

Question 37.
∫$$\frac{\sin 2x}{(a+b\cos x)^2}$$ dx on {R, if |a| > |b| I ⊂ {x ∈ R : a + b cos x ≠ x}, if |a| < |b|.
Solution:
Put a + b cos x = t ⇒ cos x = $$\frac{t-a}{b}$$
Then b(-sin x)dx = dt
⇒ sin dx = $$\frac{-1}{b}$$ dt

Question 38.
∫$$\frac{\sec x}{(\sec x+\tan x)^2}$$ dx on I ⊂ R\{(2n + 1)$$\frac{\pi}{2}$$ : n ∈ Z}.
Solution:
Put sec x + tan x = t
Then (sec x tan x + sec² x) dx = dt
sec x (sec x + tan x) dx = dt

Question 39.
∫$$\frac{dx}{a^2\sec^2 x+b^2\cos^2 x}$$ on R, a ≠ 0, b ≠ 0.
Solution:

Question 40.
∫$$\frac{dx}{\sin(x-a)\sin(x-b)}$$ on I ⊂ R$${a + nπ : n ∈ Z} ∪ {b + nπ n ∈ Z}). Solution: Question 41. ∫\(\frac{dx}{\sin(x-a)\sin(x-b)}$$ on I ⊂ R\ ({a + $$\frac{(2n+1)\pi}{2}$$ : n ∈ Z} ∪ {b + $$\frac{(2n+1)}{2}$$π : n ∈ Z}).
Solution:

III. Evaluate the following integers.

Question 1.
∫$$\frac{\sin 2x}{a \cos^2 x + b \sin^2 x}$$ dx on I ⊂ R\ {x ∈ R |a cos² x + b sin² x = 0}.
Solution:
t = a cos² x + b sin² x
dt = (a(2cos x)(-sin x) + b(2sin x cos x))dx

Question 2.
∫$$\frac{1-\tan x}{1+\tan x}$$ dx for x ∈ I ⊂ R\ {nπ – $$\frac{\pi}{4}$$ : n ∈ Z}.
Solution:

Question 3.
∫$$\frac{\cot(log x)}{x}$$ dx, x ∈ I ⊂ (0, ∞)\{e : n ∈ Z}.
Solution:
t = log x ⇒ dt = $$\frac{dx}{x}$$
∫$$\frac{1-\tan x}{1+\tan x}$$ dx = ∫cot t dt
= log(sin t) + C
= log |sin (log x)| + C

Question 4.
∫ex . cot ex dx, x ∈ I ⊂ R\ {log n π : n ∈ Z}.
Solution:
t = ex ⇒ dt = ex dx
∫ex . cot ex dx = ∫cot t dt
= log (sin t) + C
= log |sin(log x)| + C

Question 5.
∫sec (tan x) sec² x dx, on I ⊂ {x ∈ E : tan x ≠ $$\frac{(2k+1)\pi}{2}$$ for any k ∈ Z}. where E = R / {$$\frac{(2n+1)\pi}{2}$$ : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫sec(tan x) sec² x dx = ∫sec t. dt
= log tan ($$\frac{\pi}{4}+\frac{t}{2}$$ + C
= log (tan($$\frac{\pi}{4}+\frac{\tan x}{2}$$)) + C

Question 6.
∫$$\sqrt{sin x}$$cos dx on {2nπ, (2n + 1)π}, (n ∈ Z).
Solution:
t = sin x ⇒ dt = cos x dx
∫$$\sqrt{sin x}$$cos dx = ∫√t dt
= $$\frac{2}{3}$$ = t3/2 + C
= $$\frac{2}{3}$$ = (sin x)3/2 + C

Question 7.
∫tan4 x sec² x dx, x ∈ I ⊂ R\{$$\frac{(2n+1)\pi}{2}$$ : n ∈ Z}.
Solution:
t = tan x ⇒ dt = sec² x dx
∫tan4 x sec² x dx = ∫t4dt
= $$\frac{t^5}{5}$$ + C = $$\frac{(\tan x)^2}{5}$$ + C

Question 8.
∫$$\frac{2x+3}{\sqrt{x^2+3x-4}}$$ dx, x ∈ T ⊂ \{-4, 1}
Solution:
t = x² + 3x – 4
dt = (2x + 3)dx
∫$$\frac{2x+3}{\sqrt{x^2+3x-4}}$$ = ∫$$\frac{dt}{\sqrt{t}}$$
= 2√t + C
= $$\sqrt{x^2+3x-4}$$ + C

Question 9.
∫cosec² x$$\sqrt{\cot x}$$ dx on (0, $$\frac{\pi}{2}$$).
Solution:
t = cot x ⇒ dt = -cosec² x dx
∫cosec² x$$\sqrt{\cot x}$$ dx = -∫√t dt
= –$$\frac{2}{3}$$ t√t + C
= –$$\frac{2}{3}$$ cot(x)3/2 + C

Question 10.
∫sec x log(sec x + tan x)dx on (0, $$\frac{\pi}{2}$$).
Solution:
t = log(sec x + tanx)
dt = $$\frac{(\sec x.\tan x+\sec^2 x)}{(\sec x +\tan x}$$ = sec x dx
∫sec x .log(sec x + tan x)dx = ∫t dt
= $$\frac{t^2}{2}$$ + C
= $$\frac{(log(\sec x+\tan x)^2}{2}$$ + C

Question 11.
∫sin³ x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin³ x
sin³ x = $$\frac{1}{4}$$(3 sin x – sin 3x)
∫sin³ x dx = $$\frac{3}{4}$$∫sin x – $$\frac{1}{4}$$∫sin 3x dx
= –$$\frac{3}{4}$$ cos x + $$\frac{1}{12}$$ cos 3x + C
= $$\frac{1}{12}$$(cos 3x – 9 cos x) + C

Question 12.
∫cos³ x dx on R.
Solution:
cos 3x = 4 cos³ x – 3 cos x
∫cos³ x dx = $$\frac{3}{4}$$∫cos x + $$\frac{1}{4}$$∫cos 3x dx
= –$$\frac{3}{4}$$ sin x + $$\frac{1}{12}$$ sin 3x + C
= $$\frac{1}{12}$$(9 sin x + 9 sin 3x) + C

Question 13.
∫cos x cos 2x dx on R.
Solution:
cos 2x cos x = $$\frac{1}{2}$$(2cos 2x cos x)
∫cos x cos 2x dx = $$\frac{1}{2}$$∫(cos 3x + cos x) dx
= $$\frac{1}{2}$$∫cos 3x + $$\frac{1}{2}$$∫cos x
= $$\frac{1}{2}$$($$\frac{\sin 3x}{3}$$ + sin x) + C
= $$\frac{\sin 3x+3\sin x}{6}$$ + C

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos 3x cos x = $$\frac{1}{2}$$(2cos 3x . cos x)
= $$\frac{1}{2}$$(cos 4x + cos 2x)
∫cos x cos 3x dx = $$\frac{1}{2}$$∫cos 4x dx + $$\frac{1}{2}$$∫cos 2x dx
= $$\frac{1}{2}$$($$\frac{\sin 4x}{4}+\frac{\sin 2x}{2}$$ + C
= $$\frac{1}{8}$$(sin 4x + 2sin 2x) + C

Question 15.
∫cos4 x dx on R.
Solution:
cos4 x = (cos² x)² = ($$\frac{1+\cos 2x}{2}$$)²
= $$\frac{1}{4}$$ (1 + 2 cos 2x+ cos² 2x)
= $$\frac{1}{4}$$ (1 + 2 cos 2x + = $$\frac{1+\cos 4x}{2}$$)
= $$\frac{1}{8}$$ (2 + 4 cos 2x + 1 + cos 4x)
= $$\frac{1}{8}$$ (3 + 4 cos 2x + cos 4x)
= $$\frac{1}{8}$$(3∫dx + 4∫cos 2x dx + ∫cos 4x dx)
= $$\frac{1}{8}$$(3x + 4$$\frac{\sin 2x}{2}+\frac{\sin 4x}{4}$$) + C
= $$\frac{1}{32}$$(12x + 8 sin 2x + sin 4x) + C

Question 16.
∫x $$\sqrt{4x+3}$$ dx on (-$$\frac{3}{4}$$, ∞).
Solution:

Question 17.
∫$$\frac{dx}{\sqrt{a^2-(b+cx)^2}}$$ on {x ∈ R : |b + cx| < a}. where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:

Question 18.
∫$$\frac{dx}{a^2+(b+cx)^2}$$ on R, a, b, c are real numbers, c ≠ 0 and a > 0.
Solution:

Question 19.
∫$$\frac{dx}{1+e^x}$$, x ∈ R
Solution:

Question 20.
∫$$\frac{x^2}{(a+bx)^x}$$ dx, x ∈ I ⊂ R\{-$$\frac{a}{b}$$}, where a, b are real nembers, b ≠ 0.
Solution:
Put t = a + bx
dt = b dx ⇒ dx = $$\frac{1}{b}$$. dt

Question 21.
∫$$\frac{x^2}{\sqrt{1-x}}$$ dx, x ∈ (-∞, 1).
Solution: