Question

The value of the integral ${\int }_{\frac{1}{3}}^1\frac{(x-x^3{)}^{\frac{1}{3}}}{x^4}dx$

• A. $6$
• B. $0$
• C. $3$
• D. $4$

Answer 1

The correct option is A $6$

Let $I={\int }_{1/3}^1\frac{(x-x^3{)}^{\frac{1}{3}}}{x^4}dx$

Put $x=\sin \theta$

$\Rightarrow dx=\cos \theta d\theta$

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

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

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

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

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

$={\int }_{{\sin }^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}\frac{(\cos \theta {)}^{\frac{5}{3}}}{(\sin \theta {)}^{\frac{5}{3}}}cose{c}^2\theta d\theta$

$={\int }_{si{n}^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}(\cot \theta {)}^{\frac{5}{3}}cose{c}^2\theta d\theta$

Put $\cot \theta =t$

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

$\therefore I=-{\int }_{2\sqrt{2}}^0(t{)}^{\frac{5}{3}}dt$

$\Rightarrow I={\int }_0^{2\sqrt{2}}(t{)}^{\frac{5}{3}}dt$

$={\left[\frac{3}{8}\right(t{)}^{\frac{8}{3}}]}_0^{2\sqrt{2}}$

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

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

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

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

$=3\times 2$
$=6$
Hence, the correct Answer is A.