Question

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?

Answer 1

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

\[\frac{r^2}{2} [\frac{\pi}{2} -(\cos \pi - \cos 0) ] = \frac{r^2}{2} [\frac{\pi}{2} +2]\]

Answer 2

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.

Answer 3

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

Answer 4

ahahahaha how did I manage to forget to do that

Answer 5

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