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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 $\sin^2 x = 1 -\cos^2 x$.2011-04-17
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    Try to use $\sin^2 x = 1- \cos^2 x$...2011-04-17

1 Answers 1

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

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    +1, of course. I took the liberty of inserting some spacing characters to make the formulas a bit better readable.2011-04-17
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    @Theo: Thanks. I'd noticed it looked funny, but didn't think to put in the `\,`.2011-04-17