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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?
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?

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Empty
 one year ago
Best ResponseYou've already chosen the best response.0Hahaha thanks XD \[\int_0^r \frac{x^2 dx}{\sqrt{r^2x^2}}\] 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 \[\frac{r^2}{2} [\frac{\pi}{2} (\cos \pi  \cos 0) ] = \frac{r^2}{2} [\frac{\pi}{2} +2]\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahh you didn't integrate the cos before doing the limits

Empty
 one year ago
Best ResponseYou've already chosen the best response.0ahahahaha how did I manage to forget to do that

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{0}^{\pi/2}(1/2)r^2 (1cos(2\theta)d\theta\] \[\dfrac{r^2}{2}\int_{0}^{\pi/2}(1cos (2\theta))d\theta\] \[\dfrac{r^2}{2} [\theta\dfrac{sin(2\theta)}{2}]_0^{\pi/2} = \dfrac{r^2 \pi}{4}\]
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