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This question already has an answer here:

The problem is $$\int \cos^3 (x) \sin^5 (x) dx.$$

The method we've gone over in class involes splitting up the trig functions so that we can set one to U, then take du so that du matches a part of the function. But I'm struggling to find these. Any explanation is helpful.

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marked as duplicate by Hans Lundmark, Peter Taylor, beep-boop, user147263, Adam Hughes Oct 6 '14 at 18:17

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Since $$\cos^3 x\sin^5x=\cos x(1-\sin^2x)\sin^5x$$ setting $u=\sin x$ gives you $\cos xdx=du$ and

$$\int (1-\sin^2x)\sin^5x\color{red}{\cos xdx}=\int (1-u^2)u^5\color{red}{du}=\int (u^5-u^7)du.$$

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    $\begingroup$ OR $$\cos^3x\sin^5x=\sin x(1-\cos^2x)^2\cos^3x$$ $\endgroup$ – lab bhattacharjee Oct 6 '14 at 16:14
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your integrand has the following form (it is better to integrate this) $\frac{1}{128} (6 \sin (2 x)-2 \sin (4 x)-2 \sin (6 x)+\sin (8 x))$

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