Evaluation of Integral $ \int \sqrt{\sin x}\; dx$

Question

Evaluation of Integral $\displaystyle \int \sqrt{\sin x}\; dx$

$\bf{My\; Try::}$ Let $\sin x = y^2\;, $ Then $\displaystyle \cos xdx =2ydy\Rightarrow dx = \frac{2y}{\sqrt{1-y^4}}dy$

So $\displaystyle \int \sqrt{\sin x}\;dx = 2\int \frac{y^2}{\sqrt{1-y^4}}dy = 2\int y^2\cdot \left(1-y^4\right)^{-\frac{1}{2}}dy$

(Using Wolframalpha It Show The Results is in the form of Elleptical Integral of first and Second Kind.)

Now How Can I solve after that

Help Required

Thanks

Answer 1

Effectively $$\displaystyle \begin{align} & \int \sqrt{\sin x}\; dx=-2 E\left(\left.\frac{1}{4} (\pi -2 x)\right|2\right) \\ & 2\int \frac{y^2}{\sqrt{1-y^4}}dy= 2 \left(E\left(\left.\sin ^{-1}(y)\right|-1\right)-F\left(\left.\sin ^{-1}(y)\right|-1\right)\right) \end{align}$$ In my opinion, you can not get rid of the elliptic integrals. The result would be still worse using Weierstrass substitution.