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How do you prove that for all real $m$, $$\int_0^\frac{\pi}{2}\frac{\sin^mx}{\sin^mx+\cos^mx}\mathrm dx=\frac{\pi}{4}$$

I've let $y=\frac{\pi}{2}-x$ and got

$$\int_0^\frac{\pi}{2}\frac{\sin^mx}{\sin^mx+\cos^mx}\mathrm dx= \int_\frac{\pi}{2}^0\frac{\cos^my}{\cos^my+\sin^my}\mathrm dy = \int_\frac{\pi}{2}^0\frac{\cos^mx}{\cos^mx+\sin^mx}\mathrm dx $$

This is as far as I got. I know you're supposed to add the 2 results together, but the limits of integrations aren't the same. Adding both together I get,

$$\int_0^\frac{\pi}{2}\frac{\sin^mx-\cos^mx}{\sin^mx+\cos^mx}\mathrm dx$$ which doesn't simplify at all.

Thanks for all the help in advance.

Clayton
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koifish
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1 Answers1

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Your substiution is pretty close to the one needed. Consider $\frac{\pi}{2}-x=y\rightarrow dx=-dy\,$. The new bounds are $y_1=\frac{\pi}{2}-0=\frac{\pi}{2} \,$and $y_2=\frac{\pi}{2}-\frac{\pi}{2}=0$ $$I=\int^0_\frac{\pi}{2}\frac{\cos^my}{\cos^my+\sin^my}\mathrm (-dy)=\int_0^\frac{\pi}{2}\frac{\cos^my}{\cos^my+\sin^my}\mathrm dy$$ Add this with the initial integral in order to get: $$2I=\int_0^\frac{\pi}{2}\frac{\sin^mx+\cos^{m}x}{\sin^mx+\cos^mx}\mathrm dx\rightarrow I=\frac12\int_0^\frac{\pi}{2}dx=\frac{\pi}{4}$$

Zacky
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  • Okay, I actually made a mistake in my question, I meant to use that substitution. However, if I used that substitution, my limits of integration become 0 on the top and π/2 at the bottom, which would reverse the sign of and cause the numerator to become $sin^mx-cos^mx$ wouldn't it? – koifish Aug 07 '18 at 12:49
  • Let me see if I understand what are you looking for, I will edit. – Zacky Aug 07 '18 at 12:51
  • Hi I've edited my question. Perhaps it would be clearer that way. – koifish Aug 07 '18 at 12:53
  • Oh, the missunderstanding is that $\int_b^a f(x)(-dx)=\int_a^b f(x) dx$? – Zacky Aug 07 '18 at 12:54
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    I just read your edited post. Yup turns out that was the misunderstanding. Thanks. – koifish Aug 07 '18 at 12:54