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Integral of sin^5(x/2) dx
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strip out a sine and use a trig identity\[\sin^5(\frac x2)=\sin^4(\frac x2)\cdot\sin(\frac x2)=(1-\cos^2)^2\sin(\frac x2)\]\[=[1-2\cos^2(\frac x2)+\cos^4(\frac x2)]\sin(\frac x2)\]\[=\sin(\frac x2)-2\cos^2(\frac x2)\sin(\frac x2)+\cos^4(\frac x2)\sin(\frac x2)\]now this can be handled with a couple u-substitutions.
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