Indefinite integration of sin

Question

Those two integrals
$$\displaystyle\int \sqrt{\sin \left( x\right) } dx$$ and $$\int ln \left( \sin \left( x\right) \right) dx$$ one of them known if it definite. The integration is $$ \int \limits^{\pi }_{0}\sqrt{\sin \left( x\right) } dx = 2\sqrt{\frac{2}{\pi } } \Gamma \left( \frac{3}{4} \right) ^{2}$$ My question is how we can evaluate those two integrals in indefinite form? . Thanks

Answer 1

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.