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OpenStudy (anonymous):
How would you integrate pi(16sin^2(x)) dx?
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OpenStudy (mr.math):
Use \(\sin^2(x)=\large\frac{1- \cos(2x)}{2}\).
OpenStudy (anonymous):
Can I pull out the 16pi?
OpenStudy (mr.math):
Yes.
OpenStudy (anonymous):
Okay I did and I got 4pi^2..is that right?
OpenStudy (anonymous):
oh..end points are 0-pi/2
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OpenStudy (anonymous):
0 to pi/2..sorry!
OpenStudy (mr.math):
Correct!
OpenStudy (anonymous):
Thank you very much!! :))
OpenStudy (mr.math):
Glad to help.
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