TITLE: How do i evaluate this integral $ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $?
QUESTION [5 upvotes]: Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$
Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .
Thank you for any kind of help

REPLY [2 votes]: $$
\begin{align}
\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\frac{\mathrm{d}\tan(x)}{\sec^2(x)}\tag{1}\\
&=\int_1^\infty\frac{\sqrt{u}}{\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}\tag{2}\\
&=\frac12\int_0^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\tag{3}\\
&=\frac14\int_0^\infty\frac{t^{-3/4}}{(1+t)^{1/2}}\,\mathrm{d}t\tag{4}\\[6pt]
&=\tfrac14\mathrm{B}\left(\tfrac14,\tfrac14\right)\tag{5}\\[6pt]
&=\frac{\frac14\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\tag{6}\\
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)}{\sqrt\pi}\tag{7}\\
\end{align}
$$
Explanation:
$(1)$: $\mathrm{d}\tan(x)=\sec^2(x)\,\mathrm{d}x$
$(2)$: $u=\tan(x)$
$(3)$: substitution $u\mapsto\frac1u$ leaves the integral alone
$(4)$: $u^2=t$
$(5)$: definition for Beta function
$(6)$: $\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
$(7)$: $x\Gamma(x)=\Gamma(x+1)$
Note that
$$
\begin{align}
\int_{\pi/3}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\int_{\sqrt3}^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\\
&=2\int_0^{3^{-1/4}}\frac{\mathrm{d}v}{\sqrt{1+v^4}}\tag{8}
\end{align}
$$
Using the Binomial Theorem to get a series, integrating, and considering the ratios of the terms, we get
$$
\begin{align}
\int\frac1{\sqrt{1+v^4}}\,\mathrm{d}v
&=\sum_{k=0}^\infty\int\binom{2k}{k}\left(-\frac{v^4}4\right)^k\,\mathrm{d}v\\
&=\sum_{k=0}^\infty(-1)^k\frac{\binom{2k}{k}}{4^k}\frac{v^{4k+1}}{4k+1}\\
&=\vphantom{\mathrm{F}}_2\mathrm{F}_1\left(\frac14,\frac12;\frac54;-v^4\right)v\tag{9}
\end{align}
$$
At $v=3^{-1/4}$, the series above converges more than $0.477$ digits per term.
Thus, we get
$$
\begin{align}
\int_{\pi/4}^{\pi/3}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)}{\sqrt\pi}-\frac2{3^{1/4}}\,\vphantom{\mathrm{F}}_2\mathrm{F}_1\left(\frac14,\frac12;\frac54;-\frac13\right)\\[6pt]
&\doteq0.37913313649500221817\tag{10}
\end{align}
$$
This matches numerically the answer Mathematica gives, although the form is a bit different.