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
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 dθ= dt
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.