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# Choose the correct answer in the value of the integral$\int\limits_\frac{1}{3}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

$\begin{array}{1 1}6 \\ 0 \\ 3 \\ 4 \end{array}$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\frac{d}{dx}(\cot x)=- cosec ^2 x$
• (iii)$\frac{d}{dx}(\sin x)=\cos x$
Given $\int\limits_\frac{1}{3}^1\large\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$

Let $x=\sin \theta$

On differentiating we get w.r.t. x

$dx=\cos \theta d\theta$

When x=1/3, $\theta=\sin^{-1}(1/3)$ and

When x=1,$\theta=\frac{\pi}{2}$

Hence $I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}\large\frac{(\sin \theta-\sin^3 \theta)^{1/3}}{\sin ^4 \theta}\cos \theta d\theta$

Hence $I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}$$\large\frac{(\sin \theta)^{1/3} \times (1-\sin^2 \theta)^{1/3}}{\sin ^4 \theta}\cos \theta d\theta But 1-\sin^2\theta=\cos ^2\theta Hence I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}$$\large\frac{(\sin \theta)^{1/3} (\cos^2 \theta)^{1/3}}{\sin ^4 \theta}\cos \theta$
d\theta$We can split$\sin ^4 \theta\;as\;\sin ^2 \theta \times \sin^2 \theta$Hence$I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}$$\large\frac{(\sin \theta)^{1/3} (\cos^2 \theta)^{1/3}}{\sin ^2 \theta.\sin ^2 \theta}\cos \theta \theta d\theta But  \cos \theta.(\cos \theta)^{2/3}=(\cos \theta)^{5/3} and \sin^2\theta.(\sin \theta)^{1/3}=(\sin \theta)^{5/3} Hence I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}$$\large\frac{(\cos \theta)^{5/3}} {(\sin \theta)^{5/3}} \times \frac{1}{\sin ^2}$\theta}d\theta$

But $\frac{1}{\sin ^2 \theta}=cosec ^2 \theta \;and\;\frac{\cos \theta}{\sin \theta}=\cot \theta$

Hence $I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}} (\cot )^{5/3} \times cosec ^2\theta.d\theta$

Let $\cot \theta=t$

On differentiating we get

$-cosec^2\theta d\theta=dt=>cosec^2\theta d\theta=-dt$

When$\theta=\sin ^{-1}(1/3),t=0$

On substituting for t and dt

Hence $I=-\int \limits_0^{2 \sqrt 2}(t)^{5/3}.dt$

On integrating we get,

$- \bigg[\large\frac{t^{5/3+1}}{5/3+1}\bigg]_{2\sqrt 2}^0=-\bigg[\frac{t ^{8/3}}{8/3}\bigg]_{2 \sqrt 2}^0$

$=-\frac{3}{8}\bigg[t^{8/3}\bigg]_{2\sqrt 2}^0$

On applying limits,

$=\frac{3}{8}\bigg[0-(2^{3/2})^{8/3}\bigg]$

$=-\frac{3}{8}[-2^4]=\frac{3}{8} \times 16$

$=3 \times 2=6$

Hence the correct answer is A
edited Apr 8 by meena.p