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Can someone show how to get the equality $$\int_0^{\pi/2} \cos^m(x) \sin^m(x) dx = 2^{-m}\int_0^{\pi/2} \cos^m (x) dx$$ with integral substitution? Thanks!

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Hints: 1) $\sin 2x=2\sin x \cos x$ and 2) $\sin(\pi/2-y)=\cos y$. –  O.L. Aug 6 '13 at 10:30
    
@O.L.: Make it him as an answer. :) –  B. S. Aug 6 '13 at 10:57
    
Oh man, I feel dumb now...I totally forgot about trig relations and was just trying u-sub with sines and cosines. Thanks for the tip! –  paul Aug 6 '13 at 10:57
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1 Answer

HINT:

$$\int_0^{\frac\pi2}\sin^mx\cos^mxdx=\frac1{2^m} \int_0^{\frac\pi2}(\sin2x)^mdx$$

$$=\frac1{2^m} \int_0^{\pi}(\sin y)^m\frac{dy}2$$

$$=\frac1{2^m} \int_0^{\frac\pi2}(\sin y)^m\frac{dy}2+\frac1{2^m} \int_{\frac\pi2}^\pi(\sin y)^m\frac{dy}2$$

Put $y=x-\frac\pi2$ and use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\text{ to find }\int_0^{\frac\pi2}(\sin y)^mdy=\int_0^{\frac\pi2}(\cos y)^mdy$$

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