How to evaluate $\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$

Question

$$\int^{\frac{\pi}{4}}_{0}\frac{1}{(\sqrt{\sin x})}dx$$

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.

Answer 1

$$\int^{\pi}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx + \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos x}} \mathrm dx = 2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx$$

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 $y = \frac{1}{2}$

$$2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$$

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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

$$\int^{\pi/4}_{0} \frac{\mathrm dx}{\sqrt{\sin(x)}} =\frac{\Gamma \left( \frac{1}{4}\right)^2}{2\sqrt{2\pi}}-2 F\left(\frac{\pi}{8} \left. \right|2 \right)$$

Answer 2

Dividing both numerator and denominator by $\cos^2(x)$ and the substitution $\tan(x)=t$ may help.