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Question:

The value of the integral $\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is

A. 6

B. 0

C. 3

D. 4

Solution:

Let $I=\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$

Also, let $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$

When $x=\frac{1}{3}, \theta=\sin ^{-1}\left(\frac{1}{3}\right)$ and when $x=1, \theta=\frac{\pi}{2}$

$\Rightarrow I=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{\left(\sin \theta-\sin ^{3} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}\left(1-\sin ^{2} \theta\right)^{\frac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}(\cos \theta)^{\frac{2}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{(\sin \theta)^{\frac{1}{3}}(\cos \theta)^{\frac{2}{3}}}{\sin ^{2} \theta \sin ^{2} \theta} \cos \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{(\cos \theta)^{\frac{5}{3}}}{(\sin \theta)^{\frac{5}{3}}} \operatorname{cosec}^{2} \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}(\cot \theta)^{\frac{5}{3}} \operatorname{cosec}^{2} \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \frac{(\cos \theta)^{\frac{5}{3}}}{(\sin \theta)^{\frac{5}{3}}} \operatorname{cosec}^{2} \theta d \theta$

$=\int_{\sin ^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi}{2}}(\cot \theta)^{\frac{5}{3}} \operatorname{cosec}^{2} \theta d \theta$

Let cotθ = t ⇒ −cosec2θ dθdt

When $\theta=\sin ^{-1}\left(\frac{1}{3}\right), t=2 \sqrt{2}$ and when $\theta=\frac{\pi}{2}, t=0$

$\therefore I=-\int_{2 \sqrt{2}}^{0}(t)^{\frac{5}{3}} d t$

$=-\left[\frac{3}{8}(t)^{\frac{8}{3}}\right]_{2 \sqrt{2}}^{0}$

$=-\frac{3}{8}\left[(t)^{\frac{8}{3}}\right]_{2 \sqrt{2}}^{0}$

$=-\frac{3}{8}\left[-(2 \sqrt{2})^{\frac{8}{3}}\right]$

$=\frac{3}{8}\left[(\sqrt{8})^{\frac{8}{3}}\right]$

$=\frac{3}{8}\left[(8)^{\frac{4}{3}}\right]$

$=\frac{3}{8}[16]$

$=3 \times 2$

$=6$

Hence, the correct answer is A.