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JEE Main 201412 AprilOnline

Exam: JEE Main
Year: 2014
Date: 12 April
Mode: Online
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2014 · 12 April · Online · Mathematics · Indefinite Integration

The integral ∫ ( ^2 x ^2 x)/( ( ^3 x+ ^3 x)^2) d x is equal to:

\text{ The integral } \int \frac{\sin ^2 x \cos ^2 x}{

\left(\sin ^3 x+\cos ^3 x\right)^2}

d x

\text{ is equal to: }

Options

A $\frac{1}{\left(1+\cot ^3 x\right.}+c$
B $\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$
C $\frac{\sin ^3 x}{\left(1+\cos ^3 x\right.}+c$
D $-\frac{\cos ^3 x}{3\left(1+\sin ^3 x\right.}+c$

Correct answer

B. $\left.-\frac{1}{3\left(1+\tan ^3 x\right.}+c\right)$

Step-by-step solution

Let

I=\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x

\begin{aligned}&\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\sin ^3 x+\cos ^3 x}\right)^2 d x \\ &\mathrm{I}=\int\left(\frac{\sin x \cdot \cos x}{\cos ^3 x\left(1+\tan ^3 x\right)}\right)^2 d x \\ &=\int\left(\frac{\sin x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)}\right)^2 d x \\ &\text{ Put }1+\tan ^3 x=\mathrm{t}\\ &d t=3 \tan ^2 x \sec ^2 x d x \text{ or }d x \\ &=\frac{d t}{3 \tan ^2 x \sec ^2 x}\\ &\therefore \mathrm{I}=\int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{3 \tan ^2 x \sec ^2 x} \end{aligned}

\begin{aligned}&\mathrm{I}=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\frac{\sin ^2 x}{\cos ^2 x} \times \sec ^2 x}\\ &=\frac{1}{3} \int \frac{\sin ^2 x \cdot \sec ^4 x}{t^2} \times \frac{d t}{\sin ^2 x \sec ^4 x}\\ &\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2}=\frac{1}{3} \intt^{-2} d t \\ &\mathrm{I}=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c=\frac{-1}{3}\left[\frac{1}{t}\right]+c \\ &\text{ or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^3 x\right)}+c \end{aligned}

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