# Math Help - Integration

1. ## Integration

How would integrate the following?

2. Originally Posted by atwinix
How would integrate the following?

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

Let
$u=(1-x^2)^{\frac{1}{2}}$
$du =\frac{-x dx}{(1-x^2)^{\frac{1}{2}}}$
and $x^2 = 1-u^2$

so
$\int x^3 (1-x^2)^{-\frac{1}{2}}dx=\int -(1-u^2)du=\frac{u^3}{3} - u = \frac{(1-x^2)^{\frac{3}{2}}}{3} - (1-x^2)^{\frac{1}{2}}$

3. Write,
$\frac{1}{2} \int x^2 (1-x^2)^{-1/2} (2x) dx$.
Let $t=x^2$ to get,
$\frac{1}{2} \int t(1-t)^{-1/2} dt$
This is the Beta integral.

4. It's not the most efficient, but you can use trig sub. I like trig sub.

A lot of times there's an easier sub to make, but waht the heck.

$\int\frac{x^{3}}{\sqrt{1-x^{2}}}dx$

Let $x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}$

$\int\frac{sin^{3}{\theta}}{\sqrt{1-sin^{2}{\theta}}}cos{\theta}d{\theta}$

Yoiu can see the identity in the radical: $1-sin^{2}{\theta}=cos^{2}{\theta}$

So, it whittles down to:

$\int{sin^{3}{\theta}}d{\theta}=\frac{-sin^{2}{\theta}cos{\theta}}{3}-\frac{2cos{\theta}}{3}$

Now, to get it back in terms of x, sub in ${\theta}=sin^{-1}x$

You get:

$\frac{-x^{2}\sqrt{1-x^{2}}}{3}-\frac{2}{3}\sqrt{1-x^{2}}=\boxed{\frac{-(x^{2}+2)\sqrt{1-x^{2}}}{3}}$