I need help with an Integral. It is: 2(sin2x)^3... so I set u to 2x and du to 2 so this would make he integral of du(sinu)^3... now I'm stuck Thanks
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Originally Posted by alex.caine I need help with an Integral. It is: 2(sin2x)^3... so I set u to 2x and du to 2 so this would make he integral of du(sinu)^3... now I'm stuck Thanks $\displaystyle \int\sin^3u\,du$ $\displaystyle =\int\sin u\sin^2 u\,du$ $\displaystyle =\int\sin u(1-\cos^2u)\,du$ $\displaystyle =\int\sin u\,du-\int\sin u\cos^2u\,du$ Can you handle it from here?
$\displaystyle \int 2\sin^3(2x)dx = \int 2\sin(2x)(1-\cos^2(2x))$ $\displaystyle =\int2\sin(2x)-2\sin(2x)\cos^2(2x)$ $\displaystyle =-\cos(2x)-\cos^3(2x)$ (differentiate to check) Works for all odd-powered trig integrals.
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