Empty
  • Empty
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Empty
  • 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 \[\frac{r^2}{2} [\frac{\pi}{2} -(\cos \pi - \cos 0) ] = \frac{r^2}{2} [\frac{\pi}{2} +2]\]
Empty
  • 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.
dumbcow
  • dumbcow
ahh you didn't integrate the cos before doing the limits

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Empty
  • Empty
ahahahaha how did I manage to forget to do that
Loser66
  • Loser66
\[\int_{0}^{\pi/2}(1/2)r^2 (1-cos(2\theta)d\theta\] \[\dfrac{r^2}{2}\int_{0}^{\pi/2}(1-cos (2\theta))d\theta\] \[\dfrac{r^2}{2} [\theta-\dfrac{sin(2\theta)}{2}]_0^{\pi/2} = \dfrac{r^2 \pi}{4}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.