# Thread: Can you integrate this?

1. ## Can you integrate this?

I can't find how to integrate this
x^2*sqrt(2x-x^2)
I tried with part derivation but I can't make it.

2. Hello, vincisonfire!

$\displaystyle \int x^2\sqrt{2x-x^2}\,dx$

We have: .$\displaystyle 2x - x^2 \;=\;{\color{blue}1 - 1} + 2x - x^2 \;=\;1 - (x^2-2x+1) \;=\;1-(x-1)^2$

The integral becomes: .$\displaystyle \int x^2\sqrt{1-(x-1)^2}\,dx$

Let: $\displaystyle x-1 \:=\:\sin\theta \quad\Rightarrow\quad x \:=\:\sin\theta + 1 \quad\Rightarrow\quad dx \:=\:\cos\theta\,d\theta$

. . and: .$\displaystyle \sqrt{1-(x-1)^2} \;=\;\sqrt{1-\sin^2\!\theta} \;=\;\sqrt{\cos^2\!\theta} \;=\;\cos\theta$

Substitute: .$\displaystyle \int(\sin\theta+1)^2\cdot\cos\theta\cdot\cos\theta \,d\theta \;=\;\int\left(\sin^2\!\theta + 2\sin\theta + 1\right)\cos^2\!\theta\,d\theta$

. . and we have: .$\displaystyle \int\bigg(\sin^2\!\theta\cos^2\!\theta + 2\sin\theta\cos^2\!\theta + \cos^2\!\theta\bigg)\,d\theta$

And I assume you can finish it now . . .