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$$\int \sin^3{x}\,\cos^5{x}\,dx = \int \sin{x}\,(\cos^5{x}-\cos^7{x})\,dx$$

My ignorance amuses me hehe. Even if I multiply it out I still don't get it.

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How about $\sin^2 x = 1 -\cos^2 x$. – Tpofofn Apr 17 '11 at 17:32
Try to use $\sin^2 x = 1- \cos^2 x$... – Fabian Apr 17 '11 at 17:33
up vote 8 down vote accepted

$\sin^3x\,\cos^5x=\sin x\,\sin^2x\,\cos^5x=\sin x\,(1-\cos^2x)\,\cos^5x=\sin x\,(\cos^5x-\cos^7x)$.

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+1, of course. I took the liberty of inserting some spacing characters to make the formulas a bit better readable. – t.b. Apr 17 '11 at 17:40
@Theo: Thanks. I'd noticed it looked funny, but didn't think to put in the \,. – Isaac Apr 17 '11 at 17:43

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