# Please solve the attached problem in integral calculus.

Sami Ullah
46 Points
3 years ago
Its very easy just a few manipulations you’ll have to do.First I will just solve the integral and later apply the upper and lower limits.

$I=\int\frac{\sqrt{x}}{\sqrt{1-x^3}}.dx$

$Let$            $u=x^\frac{3}{2}$    $\Rightarrow$      $u^2=x^3$

$Then$       $du=\frac{3}{2}x^\frac{1}{2}.dx$

$\Rightarrow$            $\frac{2}{3}du=\sqrt{x}.dx$

$Now$  $substituting$  $values$,

$I=\int \frac{2}{3}\times \frac{1}{\sqrt{1-u^2}}.du$

$I= \frac{2}{3}\sin^{-1}(u)$

$I= \frac{2}{3}\sin^{-1}(x^\frac{3}{2})$

$Putting$ $values$,

$I= \frac{2}{3}[\sin^{-1}(1^\frac{3}{2})-\sin^{-1}(0)]$

$I= \frac{2}{3}[\frac{\pi }{2}-0]$

$I= \frac{\pi }{3}$

So the awnser is option (2).