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integral from (pi/6) to (pi/2) of cos(x)/((sinx)^8) dx
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try \(u=\sin(x), du =\cos(x)dx, u(\frac{\pi}{6})=\frac{1}{5}, u(\frac{\pi}{2})=1\)
get \[\int_{\frac{1}{2}}^1\frac{du}{u^8}\]
typo above, i meant \(u(\frac{\pi}{6})=\frac{1}{2}\)
so is the antiderivative (1/7u^7) then just plug in u(1) - u(1/2)?
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