Integration of sqrt Sin x dx

Your integral may be expressed in terms of an incomplete elliptic integral, a Legendre integral, one may prove that

$$ \int_0^x \sqrt{\sin t}\ \text dt=\sqrt{\frac{2}{\pi }} \Gamma\left(\frac{3}{4}\right)^2-2 \text{EllipticE}\left[\frac{1}{4} (\pi -2 x),2\right],\quad 0\leq x \leq \pi, $$

where $\text{EllipticE}\left[\phi,m\right]$ is a special function studied by Legendre and you will find many of its properties here.

Given $\displaystyle \int\sqrt{\sin x}\;dx$

Let $\displaystyle \sin x = t^2\Leftrightarrow \cos xdx = 2tdt\Leftrightarrow dx = \frac{2t}{\sqrt{1-t^4}}dt$

So integral convert into $\displaystyle \int t.\frac{2t}{\left(1-t^4\right)^{\frac{1}{2}}}dt$

So Integral is $\displaystyle 2\int\;t^2.\left(1-t^4\right)^{-\frac{1}{2}}dt$

Now Using $\displaystyle \bullet\; \int x^m.\left(a+bx^n\right)^p\;dx$

where $m\;,n\;,p$ are Rational no.

which is Integrable only when $\displaystyle \left(\frac{m+1}{n}\right)\in \mathbb{Z}$ or $\displaystyle \left\{\frac{m+1}{n}+p\right\}\in\mathbb{Z}$

Now here $\displaystyle 2\int\;t^2.\left(1-t^4\right)^{-\frac{1}{2}}dt$

$\displaystyle m = 2\;\;,a = 1\;\;,b = -1\;\;,n = 4\;\;,p = -\frac{1}{2}$

and $\displaystyle \left(\frac{2+1}{4}\right)\neq \mathbb{Z}$ or $\displaystyle \left(\frac{2+1}{4}\right)-\frac{1}{2}\neq \mathbb{Z}$

So We can not integrate $\displaystyle \int\sqrt{\sin x}\;dx =2\int\;t^2.\left(1-t^4\right)^{-\frac{1}{2}}dt$ in terms of elementry function.

Let $u=\sqrt{\sin x}$ ,

Then $x=\sin^{-1}u^2$

$dx=\dfrac{2u}{\sqrt{1-u^4}}du$

$\therefore\int\sqrt{\sin x}~dx$

$=\int\dfrac{2u^2}{\sqrt{1-u^4}}du$

$=\int2u^2\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n}}{4^n(n!)^2}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n+2}}{2^{2n-1}(n!)^2}du$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!u^{4n+3}}{2^{2n-1}(n!)^2(4n+3)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!\sin^{2n+\frac{3}{2}}x}{2^{2n-1}(n!)^2(4n+3)}+C$