Is there any (elementary) way to evaluate the integral∫0π2(cos⁡x)a(1−cos⁡x)1−adxfor any a∈(0,1)?

∫0π2(cos⁡x)a(1−cos⁡x)1−adx=∫01xa(1−x)1−a1−x2dx=∫01xa(1−x)12−a(1+x)−12dxThe following integral representation for 2F1:2F1(α,β;γ;z)=Γ(γ)Γ(β)Γ(γ−β)∫01xβ−1(1−x)γ−β−1(1−zx)−αdxgives∫0π2(cos⁡x)a(1−cos⁡x)1−adx=Γ(a+1)Γ(32−a)Γ(52)2F1(12,a+1;52;−1)

Step by step solution to "Is there any (elementary) way to evaluate the..."

College
Calculus and Analysis
Integral Calculus
Applications of integrals

Jefferson Pacheco

Answered question

2022-04-13

Is there any (elementary) way to evaluate the integral

\int_{0}^{\frac{π}{2}} {(\cos x)}^{a} {(1 - \cos x)}^{1 - a} d x

for any

a \in (0, 1)

Answer & Explanation

zvonkurm7h

Beginner2022-04-14Added 8 answers

\int_{0}^{\frac{π}{2}} {(\cos x)}^{a} {(1 - \cos x)}^{1 - a} d x = \int_{0}^{1} \frac{x^{a} {(1 - x)}^{1 - a}}{\sqrt{1 - x^{2}}} d x = \int_{0}^{1} x^{a} {(1 - x)}^{\frac{1}{2} - a} {(1 + x)}^{- \frac{1}{2}} d x

The following integral representation for

_{2} F_{1}

_{2} F_{1} (α, β; γ; z) = \frac{Γ (γ)}{Γ (β) Γ (γ - β)} \int_{0}^{1} x^{β - 1} {(1 - x)}^{γ - β - 1} {(1 - z x)}^{- α} d x

gives

\int_{0}^{\frac{π}{2}} {(\cos x)}^{a} {(1 - \cos x)}^{1 - a} d x = {\frac{Γ (a + 1) Γ (\frac{3}{2} - a)}{Γ (\frac{5}{2})}}_{2} F_{1} (\frac{1}{2}, a + 1; \frac{5}{2}; - 1)

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