Question

integrate (sin x)^6 please

Answer 1

so, using the identities of derivatives, \[{d \over dx} (\sin x)^6 = 6(\sin x)^5 \cos x\]

Answer 2

@KingGeorge, he asked to integrate it

Answer 3

xD

Answer 4

Here's the method to solve it. Rewrite it slightly as \[\sin^6 x=\sin^4x * \sin^2x=\sin^4x(1-\cos^2x)\] Use an integration by parts with \[u=\sin^4x, v=1-\cos^2x\] This will give you something known as the sine reduction formula, this will decrease the power of sin in your integral from 6 to 4. Do it again to get from 4 to 2. Then rewrite \[\sin^2x=\frac{1-\cos(2x)}{2}\] to solve the final part of your integral. It is very messy, but purely computational and not too hard once you know the trick.

Answer 5

\[\int sin^n(x)=-\frac{1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\int sin^{n-2}(x)dx\]