Find $\int\frac{x^2}{\sqrt{1-x^2}}$ I need to compute the following integral:
$$\int\frac{x^2}{\sqrt{1-x^2}}~dx$$
I tried everything I know, including defining $u=x^2$ and $u=\sqrt{1-x^2}$, but it failed.
 A: Hint: Try the trigonometric substitution $x=\sin\theta.$ Keep the Pythagorean identity in mind.
A: Let $x = \sin(u)$, then we have $dx = \cos(u) du$
Substituting in above, we get:
$$\int\frac{x^2}{\sqrt{1 - x^2}} dx = \int \frac{\sin^2(u)}{\sqrt{1 - \sin^2(u)}} \cdot \cos(u) du = \int\frac{\sin^2(u)}{\cos(u)} \cdot \cos(u) du = \int \sin^2(u) du$$
which solves to:
$$\frac{1}{2} \cdot (u - \sin(u) \cos(u))$$
Now you can make the reverse substitution to get the desired result, which is:
$$\frac{1}{2} \cdot \left(\sin^{-1}(x) - \sin(\sin^{-1}(x)) \cos(\sin^{-1}(x))\right) = \frac{1}{2}\cdot \left(\sin^{-1}(x) - \frac{x}{\sqrt{1 - x^2}}\right)$$
which is the required answer. That is:
$$\color{red}{\int\frac{x^2}{\sqrt{1 - x^2}} dx = \frac{1}{2}\cdot \left(\sin^{-1}(x) - \frac{x}{\sqrt{1 - x^2}}\right)}$$
NOTE: The last step is due to the fact that:
$$\sin(\sin^{-1}(x)) = x\,, \textrm{and}\, \cos(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}$$
A: hint
$$\frac{x^2}{\sqrt{1-x^2}}=\frac{-(1-x^2)+1}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}-\sqrt{1-x^2}$$
A: Substitute $x=\sin \theta$,
$$\mathrm dx=\cos \theta\mathrm d\theta$$
the integral becomes
$$\int \frac{\sin^2 \theta .\cos \theta }{\cos \theta}\mathrm d\theta.$$
