<s>[INST] How do I evaluate $ \int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x, \textrm{ where }0 \leq a\leq 1$? We tackle the integral by converting it into a Beta function by letting $y=\sin^2x$, then
$$
\begin{aligned}
 \int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x &=\int_{0}^{1} \frac{y^{\frac{a}{2}}  d y}{2 y^{\frac{1}{2}}(1-y)^{\frac{1}{2}}} \\
&=\frac{1}{2} \int_{0}^{1}y^ \frac{a-1}{2}(1-y)^{-\frac{1}{2}} d y \\
&=\boxed{\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right)}
\end{aligned}
$$
Also it can be expressed in terms of Gamma function by the property of Beta function.
$$
\begin{aligned}
 \int_{0}^{\frac{\pi}{2}} \sin ^{a} x d x &=\frac{\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{a}{2}+1\right)} \\
&=\boxed{\frac{\sqrt{\pi} \Gamma\left(\frac{a+1}{2}\right)}{a \Gamma\left(\frac{a}{2}\right)}}
\end{aligned}
$$
For example, $$
\begin{aligned}
&\int_{0}^{\frac{\pi}{2}} \sin x d x=\frac{1}{2} B\left(1, \frac{1}{2}\right)=1 \\
&\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} d x=\frac{2 \sqrt{\pi} \left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\\& \int_{0}^{\frac{\pi}{2}} \sqrt[3]{\sin x} d x=\frac{3 \sqrt{\pi} \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)}\\&\int_{0}^{\frac{\pi}{2}} \sqrt[6]{\sin x} d x=\frac{1}{2} B\left(\frac{7}{12}, \frac{1}{2}\right)
\end{aligned}
$$
My Question
Is there any other closed form of the integral?
 [/INST] First, there is an antiderivative : for $0\leq x \leq \frac \pi 2$ $$I=\int \sin ^{a}( x)\, dx=-\cos (x) \,\,
   _2F_1\left(\frac{1}{2},\frac{1-a}{2};\frac{3}{2};\cos ^2(x)\right)$$ In terms of summation
$$I=-\frac{1}{\Gamma \left(\frac{1-a}{2}\right)}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1-a}{2}\right)}{(2 n+1)\, \Gamma (n+1)}\,\cos ^{2 n+1}(x)$$
</s>