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integral from pi/2 to 0 sin^5x dx
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Backwards? O.o not 0 to pi/2?
No pi/2 to 0
I suppose it doesnt matter, same area covered. You need to break sin^5(x) up like this: \[(\sin ^{2}x)(\sin ^{2}x)(sinx)\] After that, use the pythagorean identity of: \[\sin ^{2}x + \cos ^{2}x = 1\]to change those two sin^2(x)s into cosines. Then you can integrate like normal.
|dw:1377843881848:dw|
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