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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} $

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1 Answer

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  • (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}}$$\large\frac{(\sin \theta)^{1/3} \times (1-\sin^2 \theta)^{1/3}}{\sin ^4 \theta}\cos \theta$
But $1-\sin^2\theta=\cos ^2\theta$
Hence $I=\int \limits_{\sin^{-1}(1/3)}^{\frac{\pi}{2}}$$\large\frac{(\sin \theta)^{1/3} (\cos^2 \theta)^{1/3}}{\sin ^4 \theta}\cos \theta$
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}$
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,
$=-\frac{3}{8}[-2^4]=\frac{3}{8} \times 16$
$=3 \times 2=6$
Hence the correct answer is A
answered Feb 13, 2013 by meena.p
edited Apr 8 by meena.p
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