My Beta Functions - Part 1

Calculus Level 4

∫ 0 π / 2 cos ⁡ 2 m θ sin ⁡ 2 n θ d θ = ? \large \displaystyle \int_{0}^{{\pi} / {2}} \cos^{2m}\theta \sin^{2n}\theta \ d\theta = \, ?

1 2 β ( m + 1 2 , n + 1 2 ) \frac{1}{2} \beta(m + \frac{1}{2},n + \frac{1}{2} ) None of these options β ( 2 m + 1 , 2 n + 1 ) \beta(2m+1,2n+1) 2 β ( 2 m , 2 n ) 2 \beta(2m,2n)

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Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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2 solutions

Let I = ∫ 0 π / 2 cos ⁡ 2 m θ sin ⁡ 2 n θ d θ \large \displaystyle \int_{0}^{{\pi} / {2}} \cos^{2m}\theta \sin^{2n}\theta d\theta T a k e s i n θ = t Take sin\theta = t S o , c o s θ d θ = d t So, cos\theta d\theta = dt

Limits change as 0 to 1.

I = ∫ 0 1 t 2 n ( 1 − t 2 ) 2 m − 1 2 d t \large \displaystyle \int_{0}^{1} t^{2n} (1 - t^{2})^\frac{2m-1}{2} dt

Again Take t^{2} = k ,

2 t d t = d k 2t dt = dk

d t = d k 2 ( k ) dt = \frac{dk}{2\sqrt(k)}

I = 1 2 ∫ 0 1 k n ( k ) ( 1 − k ) 2 m − 1 2 d k \frac{1}{2} \large \displaystyle \int_{0}^{1} \frac{k^{n}}{\sqrt(k)} (1 - k)^\frac{2m-1}{2} dk = 1 2 ∫ 0 1 k n + 1 2 − 1 ( 1 − k ) m + 1 2 − 1 d k \frac{1}{2} \large \displaystyle \int_{0}^{1} k^{n + \frac{1}{2} - 1} (1 - k)^{m + \frac{1}{2} - 1} dk

Here we see from the definition of Beta function that , ∫ 0 1 x m − 1 ( 1 − x ) n − 1 d t = β ( m , n ) \large \displaystyle \int_{0}^{1} x^{m - 1} (1 - x)^{n-1} dt = β(m,n) Comparing I with the above form we derive, I = 1 2 β ( m + 1 2 , n + 1 2 ) \frac{1}{2}β(m + \frac{1}{2},n + \frac{1}{2})

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Moderator note:

Simple standard approach.

The first 2 steps could be combined by taking the substitution sin ⁡ 2 θ = k \sin^2 \theta = k , though that could introduce complications in the change of variables if one is not careful.

Just compare with the standard definition of b e t a beta functions i.e. β ( x , y ) = ∫ 0 1 t x − 1 ( 1 − t ) y − 1 d t \beta (x,y)=\int _{ 0 }^{ 1 }{ { t }^{ x-1 } } { (1-t) }^{ y-1 }dt

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