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  • 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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  1. Empty
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ahh you didn't integrate the cos before doing the limits

  4. Empty
    • one year ago
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    ahahahaha how did I manage to forget to do that

  5. Loser66
    • one year ago
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    \[\int_{0}^{\pi/2}(1/2)r^2 (1-cos(2\theta)d\theta\] \[\dfrac{r^2}{2}\int_{0}^{\pi/2}(1-cos (2\theta))d\theta\] \[\dfrac{r^2}{2} [\theta-\dfrac{sin(2\theta)}{2}]_0^{\pi/2} = \dfrac{r^2 \pi}{4}\]

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