Question

Integrate :
$\int \frac{x^{2}si{n}^{-1}x}{(1-x^2{)}^{\frac{5}{2}}}$

Answer 1

$\int \frac{x^2{\sin }^{-1}x}{{(1-x^2)}^{5/2}}dx$

let ${\sin }^{-1}x=t$ ; $x=sint$

$\frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}$

$dt=\frac{1}{\sqrt{1-x^2}}dx$

$\therefore$ $\int \frac{x^2{\sin }^{-1}x}{{(1-x^2)}^{1/2}{(1-x^2)}^2}dx$

$=\int \frac{x^2}{{(1-x^2)}^2}tdt$

$=\int \frac{{\sin }^{2}t}{{(1-{\sin }^{2}t)}^2}tdt$

$=\int \frac{{\sin }^{2}t}{{\left({\cos }^{2}t\right)}^2}tdt$

$=\int {\tan }^{2}t.{\sec }^{2}ttdt$

$=\int {\tan }^{2}t{\sec }^{2}tdtt-\int \left((\int {\tan }^{2}t{\sec }^{2}tdt)\frac{d}{dt}\left(t\right)\right)dt$

$=\frac{{\tan }^{3}t}{3}.t-\int \frac{{\tan }^{3}t}{3}dt$

$=t\frac{{\tan }^{3}t}{3}-\frac{1}{3}[\int {\tan }^{2}ttantdt]$

$=t\frac{{\tan }^{3}t}{3}-\frac{1}{3}[\int {\sec }^{2}t.tantdt-\int tantdt]$

$=t\frac{{\tan }^{3}t}{3}-\frac{1}{3}\left(\frac{{\tan }^{2}t}{2}\right)-\frac{1}{3}ln\left(sect\right)+c$

$=\left({\sin }^{-1}x\right)\frac{{\tan }^3\left({\sin }^{-1}x\right)}{3}-\frac{{\tan }^2\left({\sin }^{-1}x\right)}{6}-\frac{ln\left(\sec \left({\sin }^{-1}x\right)\right)}{3}+c$