how can i integrate $\frac{x^2}{2\sqrt{1-x^2}}$

Question

$\frac{x^2}{2\sqrt{1-x^2}}$

I first thought about splitting the function this way:

=$\frac{x^2-1+1}{2\sqrt{1-x^2}}$
=$\frac{-(1-x^2)+1}{2\sqrt{1-x^2}}$ =$\frac{-(1-x^2)}{2\sqrt{1-x^2}}+\frac{1}{2\sqrt{1-x^2}}$ =${-\frac{\sqrt{1-x^2}}{2}}+\frac{1}{2\sqrt{1-x^2}}$

Now I know how to integrate the 2nd term but how do I integrate the first term?

Answer 1

If you do a trig sub you get that $\cos(\theta)=\sqrt{1-x^2},$ $\sin(\theta)=x,$ and $dx= \cos(\theta)d\theta,$ so the integral becomes

$$\int \frac{x^2}{2\sqrt{1-x^2}}dx=\int \frac{\sin^2(\theta)}{2\cos(\theta)}\cos(\theta)d\theta.$$

Integrating $\sin^2(\theta)/2$ produces the desired result after substituting back to get a function of $x.$

So you get $$\int \sin^2(\theta)d\theta=\theta/4-\sin(2\theta)/8+C$$ so

$$\int \frac{x^2}{2\sqrt{1-x^2}}dx=\arcsin(x)/4-x\sqrt{1-x^2}/4+C$$

Answer 2

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

$du=\frac{-x}{\sqrt{1-x^2}}dx$ ⇒ $u=\sqrt{1-x^2}$

$v=-(1/2)x ⇒ dv=-(1/2)dx$

⇒ $I=-(1/2)x\sqrt{1-x^2}+(1/2)\int\sqrt{1-x^2}dx$

Now take $x= \sin t$ and continue.