Question

Solve $\int \frac{x^2}{\sqrt{1-x^2}}dx$

Answer 1

Consider the given integral.

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

$I=\int \frac{x^2-1+1}{\sqrt{1-x^2}}dx$

$I=\int \frac{x^2-1}{\sqrt{1-x^2}}dx+\int \frac{1}{\sqrt{1-x^2}}dx$

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

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

We know that

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+C$

$\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\left(\frac{x}{a}\right)+C$

Therefore,

$I={\sin }^{-1}x-[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(x\right)]+C$

$I=\frac{1}{2}{\sin }^{-1}x-\frac{1}{2}x\sqrt{1-x^2}+C$

Hence, this is the answer.