Integration of sqrt Sin x dx

$$\int \sqrt{\sin x}\ \text dx$$

I asked my teachers and they said that this integration is pretty next level and will be taught in college.
Can anyone help?

Solutions Collecting From Web of "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 $x=\sin^{-1}u^2$

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

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

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

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

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

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

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

Integral of root sin x cannot exactly be found out because it has a sharp change of slope. For a sharp change, integration cannot be done.