Definite Integral $\sqrt{\sin x}$: Answers

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SUMMARY

The discussion focuses on the evaluation of the definite integrals $\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx$ and $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}dx$. Utilizing the properties of the Beta and Gamma functions, the first integral is determined to be $(1/2) \cdot \text{Beta}(3/4, 1/2)$, while the second integral equals $(1/2) \cdot \text{Beta}(1/4, 1/2)$. The product of these two integrals results in the value of $\pi$.

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$\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx.\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}dx =$
 
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Knowing the basic properties of the functions Beta and Gamma, it is easy to obtain :
first integral = (1/2)*Beta(3/4 ; 1/2)
second integral = (1/2)*Beta(1/4 ; 1/2)
and the product of the integrals = pi.
 

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