## Empty one year ago For some reason I am messing up this integral and I can't seem to get the same answer as wolfram alpha, can someone point out my error?

1. Empty

Hahaha thanks XD $\int_0^r \frac{x^2 dx}{\sqrt{r^2-x^2}}$ So first I substitute in $x=r \sin \theta$$dx = r \cos \theta d \theta$ and get $\int_0^{\pi/2} r^2 \sin^2 \theta d \theta$ Which then pull out r since it's constant and replace with the identity $2\sin^2 \theta = 1- \cos 2 \theta$ Then I end up with $\frac{r^2}{2} [\frac{\pi}{2} -(\cos \pi - \cos 0) ] = \frac{r^2}{2} [\frac{\pi}{2} +2]$

2. Empty

The answer should be $$\frac{r^2 \pi}{4}$$ so somehow I've messed up this extra cosine term and it should cancel itself out... but I guess I'm too braindead to realize what's wrong with it lol.

3. dumbcow

ahh you didn't integrate the cos before doing the limits

4. Empty

ahahahaha how did I manage to forget to do that

5. Loser66

$\int_{0}^{\pi/2}(1/2)r^2 (1-cos(2\theta)d\theta$ $\dfrac{r^2}{2}\int_{0}^{\pi/2}(1-cos (2\theta))d\theta$ $\dfrac{r^2}{2} [\theta-\dfrac{sin(2\theta)}{2}]_0^{\pi/2} = \dfrac{r^2 \pi}{4}$