The integral ∫ ^ 2 x ^ 2 x ( ^ 5 x + ^ 3 x ^ 2 x + ^ 3 x ^ 2 x + ^ 5 x) ^ 2 dx , is equal to (where C is the constant of integration).

Correct answer: C. - 1 3(1 + ^ 3 x) + C ∫ ^ 2 x ^ 2 x ( ^ 5 x + ^ 3 x ^ 2 x + ^ 3 x ^ 2 x + ^ 5 x) ^ 2 dx = ∫ ^ 2 x ^ 2 x ( ^ 3 x( ^ 2 x + ^ 2 x) + ^ 3 x( ^ 2 x + ^ 2 x)) ^ 2 dx = ∫ ^ 2 x ^ 2 x ( ^ 3 x + ^ 3 x) ^ 2 dx = ∫ ^ 2 x ^ 2 x ( ^ 3 x) ^ 2 ( ^ 3 x + 1) ^ 2 dx = ∫ ^ 2 x sec ^ 2 x ( ^ 3 x + 1) ^ 2 dx Put, ^ 3 x + 1 = t ⇒ dt = 3 ^ 2 x sec ^ 2 xdx = ∫ dt 3 (t) ^ 2 = (- 1)/(3t) + C , where C is the constant of integration. = - 1 3( ^ 3 x + 1) + C

JEE Main 20188 April

Exam: JEE Main
Year: 2018
Date: 8 April
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2018 · 8 April · Mathematics · Indefinite Integration

The integral ∫ ^ 2 x ^ 2 x ( ^ 5 x + ^ 3 x ^ 2 x + ^ 3 x ^ 2 x + ^...

The integral

∫ \dfrac{\sin^{2}x\cos^{2}x}{{\left(\sin^{5}x + \cos^{3}x\sin^{2}x + \sin^{3}x\cos^{2}x + \cos^{5}x\right)}^{2}}dx

, is equal to

(where

C

is the constant of integration).

Options

A $\dfrac{- 1}{1 + \cot^{3}x} + C$
B $\dfrac{1}{3\left(1 + \tan^{3}x\right)} + C$
C $\dfrac{- 1}{3\left(1 + \tan^{3}x\right)} + C$
D $\dfrac{1}{1 + \cot^{3}x} + C$

Correct answer

C. $\dfrac{- 1}{3\left(1 + \tan^{3}x\right)} + C$

Step-by-step solution

∫ \dfrac{\sin^{2}x\cos^{2}x}{{\left(\sin^{5}x + \cos^{3}x\sin^{2}x + \sin^{3}x\cos^{2}x + \cos^{5}x\right)}^{2}}dx

= ∫ \dfrac{\sin^{2}x\cos^{2}x}{{\left(\sin^{3}x\left(\sin^{2}x + \cos^{2}x\right) + \cos^{3}x\left(\cos^{2}x + \sin^{2}x\right)\right)}^{2}}dx

= ∫ \dfrac{\sin^{2}x\cos^{2}x}{{\left(\sin^{3}x + \cos^{3}x\right)}^{2}}dx

= ∫ \dfrac{\sin^{2}x\cos^{2}x}{{\left(\cos^{3}x\right)}^{2}{\left(\tan^{3}x + 1\right)}^{2}}dx

= ∫ \dfrac{\tan^{2}x{sec}^{2}x}{{\left(\tan^{3}x + 1\right)}^{2}}dx

Put,

\tan^{3}x + 1 = t

⇒ dt = 3\tan^{2}x{sec}^{2}xdx

= ∫ \dfrac{dt}{3{\left(t\right)}^{2}}

= \dfrac{- 1}{3t} + C

, where

C

is the constant of integration.

= \dfrac{- 1}{3\left(\tan^{3}x + 1\right)} + C

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