How to evaluate $\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$ $$\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$$
I tried everything substitution,beta gamma function everything but I am unable to convert this into a standard form so How do I solve this problem.
 A: $$\int^{\pi}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx + \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos x}} \mathrm dx = 2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx$$
Then use the formula 
$$B(x,y) = 2\int^{\pi/2}_0 (\sin t)^{2x-1} (\cos t)^{2y-1} dt$$
For $2y-1 = 0$ and $2x-1 = -\frac{1}{2}$ we have $x = \frac{1}{4}$ and $y = \frac{1}{2}$
$$2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$$

Remark
It seems the question was edited 
Define the incomplete elliptic integral of the first kind as 
$$F(\phi,k) = \int^{\phi}_0 \frac{\mathrm dx}{\sqrt{1-k^2\sin^2 x}}$$
Suppose that $0 < \phi \leq \frac{\pi}{4}$ and $k=\sqrt{2}$
$$F(\phi,\sqrt{2}) = \int^{\phi}_0 \frac{\mathrm dx}{\sqrt{\cos(2x)}} = \frac{1}{2}\int^{\pi/2}_{\pi/2-2\phi} \frac{\mathrm dx}{\sqrt{\sin(x)}}  $$
$$F(\phi,\sqrt{2})= \frac{1}{2}\int^{\pi/2}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}}-\frac{1}{2}\int^{\pi/2-2\phi}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}}$$
Hence we have 
$$\int^{\pi/2-2\phi}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F(\phi,\sqrt{2})$$
For the case $\phi = \pi/8$
$$\int^{\pi/4}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F(\pi/8,\sqrt{2})$$
Or using the other representation 
$$\int^{\pi/4}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F\left(\frac{\pi}{8} \left. \right|2 \right)$$
A: Dividing both numerator and denominator by $\cos^2(x)$ and the substitution $\tan(x)=t$ may help.
