# Math Help - substitution

1. ## substitution

Thanks!

2. Originally Posted by gralla55
Thanks!
Is this

$\int{2\pi x \sqrt{4 - x^2}\,dx}$?

3. Yep!

4. $\int{2\pi x \sqrt{4 - x^2}\,dx} = \int{2\pi x (4 - x^2)^{\frac{1}{2}}\,dx}$.

Let $u = 4 - x^2$ so that $\frac{du}{dx} = -2x$.

You can rewrite your integral as

$-\pi \int{ (4 - x^2)^{\frac{1}{2}}(-2x)\,dx}$

$= -\pi\int{u^{\frac{1}{2}}\,\frac{du}{dx}\,dx}$

$= -\pi \int{u^{\frac{1}{2}}\,du}$

$= -\frac{2\pi}{3}u^{\frac{3}{2}} + C$

$= C -\frac{2\pi \sqrt{(4 - x^2)^3}}{3}$.