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The beta function is

$$B(x,y) = 2\int_{0}^{\frac{\pi}{2}} \cos^{2x-1}\theta \sin^{2y-1}\theta d\theta$$

for positive values of $x$ and $y$.

How can this integral be used to evaluate other limits of integration?

For example,

$$ \int_{0}^{\pi} \sqrt{\sin\theta} \, d\theta$$

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    $\begingroup$ In this case, $\sin (\pi - \theta) = \sin \theta$ helps. $\endgroup$ Oct 25, 2016 at 15:10

2 Answers 2

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Consider: \begin{align} \int_{0}^{\pi} \sqrt{\sin\theta} \, d\theta &= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{\pi/2}^{\pi} \sqrt{\sin\theta} \, d\theta \\ &= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{0}^{\pi/2} \sqrt{\sin(\theta + \pi/2)} \, d\theta \\ &= \int_{0}^{\pi/2} \sqrt{\sin\theta} \, d\theta + \int_{0}^{\pi/2} \sqrt{\cos\theta} \, d\theta \\ &= \frac{1}{2} \, \left( B\left(\frac{3}{4}, \frac{1}{2} \right) + B\left( \frac{1}{2} , \frac{3}{4}\right) \right) \\ &= B\left(\frac{3}{4}, \frac{1}{2} \right) = \frac{4\pi}{s}, \end{align} where $s$ is the lemniscate constant.

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$B(\frac{1}{2},\frac{3}{4})=2\int_0^{\frac{\pi}{2}} \sqrt{\sin(\theta)} d\theta=\int_0^{\pi} \sqrt{\sin(\theta)}d\theta$
(see Daniel comment for the last step)
More generally, special values for x an y yield specific powers of the trigonometric functions

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