Question

Solve: $\int \frac{x\cdot {\sin }^{-1}x}{(1-x^2{)}^{\frac{3}{2}}}dx$.

Answer 1

$\int \frac{x{\sin }^{-1}x}{(1-x^2{)}^{3/2}}dx$

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

Let ${\sin }^{-1}x=t$

$\frac{dx}{(1-x^2{)}^{1/2}}=dt$

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

$=\int t\frac{\sin t}{{\cos }^{2}t}dt$

$=\int t\tan t\sec tdt$

$=t\int \tan t\sec tdt-\int \left[\left(\frac{dt}{dt}\right)\int \left(\sec t\tan tdt\right)\right]dt$.

$=t\sec t-\int 1\sec tdt$

$=t\sec t-log|\sec t+\tan t|+c$

$={\sin }^{-1}x\sec \left({\sin }^{-1}x\right)-log|\sec \left({\sin }^{-1}x\right)+\tan \left({\sin }^{-1}x\right)|+c$

$\therefore \int \frac{x{\sin }^{-1}xdx}{(1-x^2{)}^{3/2}}={\sin }^{-1}x\sec \left({\sin }^{-1}x\right)-log|\sec \left({\sin }^{-1}x\right)+\tan \left({\sin }^{-1}x\right)|+c$.