# Integration, trigonometry, gamma/beta functions [closed]

In a classical mechanics text, I saw an integration(as part of a math) that uses gamma function as follows:

$$\int_0^{\pi/2} (\cos^2 x-\cos^3 x)dx$$ $$=\frac{\Gamma(\frac{2+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{2+0+2}{2})}-\frac{\Gamma(\frac{3+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{3+0+2}{2})}$$ What is the exact formula that is used here?

## closed as unclear what you're asking by Felix Marin, haqnatural, Ali Caglayan, Parcly Taxel, John BOct 24 '16 at 10:13

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They are using the Beta function and in particular its defining property: $$\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=B(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,\mathrm{d}\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0$$

• I really wonder why they make things so complex ! Cheers. – Claude Leibovici Oct 23 '16 at 8:59
• Indeed. Although I admit that it's quite an easy formula to remember, and I've always found it difficult to recall the exact values of the integrals of the powers of $\sin$ and $\cos$, and always end up having to calculate them from scratch – b00n heT Oct 23 '16 at 10:03
• Can someone guide me into the steps? – Pragyaditya Das Jul 29 '17 at 19:47
• Got it. Thanks. – Pragyaditya Das Jul 29 '17 at 20:08