Intergral help

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I'm having trouble figuring out this indefinite integral.

its intergral sin^32xcos2xdx

cosxdx=dsinx, i.e. can get it to be sin^3xd(sin(x))=u^3du
==>2sin^4cos^4

Just FYI, you can use Wolfram Alpha to solve integrals, but it's good to know how to solve integrals by hand if you're not in front of a computer or if you're taking an exam...

What you've written isn't very clear. I don't know if you mean sin(x) to the power of 32 or the cube of sin(2x). Assuming that you mean the cube of sin(2x), use the substitution:

u = sin(2x),

then we get:

du = 2*cos(2x),

so the integral under this substitution reduces to:

(1/2)*integral u^3 du = (1/8 )*(u^4),

Substituting u = sin(2x) back into the formula, we have:

integral ((sin(2x))^3)*cos(2x) dx = (1/8 )*(sin(2x))^4