2
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – mrnovice Mar 25 '17 at 4:00
  • 1
    $\begingroup$ I am not sure what happened here but it seems someone edited the question wrongly and changed the integration limits. $\endgroup$ – Zaid Alyafeai Mar 25 '17 at 5:45
6
$\begingroup$

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

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

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

$\endgroup$
  • 1
    $\begingroup$ And what is $\Gamma(1/4)$? $\endgroup$ – Mark Viola Mar 25 '17 at 4:16
  • $\begingroup$ @Dr.MV, some transcendental number. $\endgroup$ – Zaid Alyafeai Mar 25 '17 at 4:21
  • $\begingroup$ Yes, what is gained over writing $\int_0^\pi \frac{1}{\sqrt{\sin(x)}}\,dx$ $\endgroup$ – Mark Viola Mar 25 '17 at 4:26
  • $\begingroup$ @Dr.MV, this shows that the integral can't be simplified to a simpler representation at least in terms of elementary functions. $\endgroup$ – Zaid Alyafeai Mar 25 '17 at 4:32
  • $\begingroup$ Well, $\Gamma$ is not an elementary function. We can also write this in terms of the Complete Elliptic Integral of the First Kind. Anyway (+1) $\endgroup$ – Mark Viola Mar 25 '17 at 4:38
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.