# Math Help - integration of x^3*sqrt(90x^2)dx using sub x=sin(u)

1. ## integration of x^3*sqrt(90x^2)dx using sub x=sin(u)

Question states using the substitution x=3 sin(x) the integral $\int {x^3 \sqrt {9 - x^2 } dx}$ becomes?

Answer is given as $
3^4 \int {\sin ^3 (u) - \sin ^5 (u)} du
$

I don't understand how to work this out

2. I think you're missing a 3 there: ${\color{red}x = 3\sin u} \ \Rightarrow \ {\color{blue}dx = 3\cos u \ du}$

So: $\int {\color{red}x}^3 \sqrt{9-{\color{red}x}^2} \ {\color{blue}dx} = \int ({\color{red}3 \sin u})^3\sqrt{9-({\color{red}3\sin u})^2} \ {\color{blue}3 \cos u \ du}$
$= \int 3^3 \sin^3 u \sqrt{9\left(1 -\sin^2 u\right)} 3\cos u \ du = 3^5 \int \sin^3 u \cos^2 u \ du$

Convert $\cos^2 u$ into terms of $\sin u$ and you should be fine.