Show two integrals are equal

**StaryNight** · April 15th 2011, 12:56 AM

How do I show that

$\int_{0}^{\pi/2} sin(2x)^{2n-1} dx = \int_{0}^{\pi/2} sin(x)^{2n-1} dx$

I've tried the substitution u = 2theta but this ends up changing the limits. I suspect I have to make an argument from symmetry?

**mr fantastic** · April 15th 2011, 01:08 AM

Originally Posted by StaryNight

How do I show that

$\int_{0}^{\pi/2} sin(2x)^{2n-1} dx = \int_{0}^{\pi/2} sin(x)^{2n-1} dx$

I've tried the substitution u = 2theta but this ends up changing the limits. I suspect I have to make an argument from symmetry?

Yes, make the substitution and then use a symmetry argument.

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