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
This is not an answer but it is too long for a comment.
As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos ^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi }}-\sqrt{2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{1}{4}\right)\approx 0.379133$$ which does not match with Mercy's result $$\ln\left(\frac{1+\sqrt2}{\sqrt3}\right) \approx 0.332067$$
Reusing all steps done for Mercy, I thing that in fact, we should arrive to $$I=\int_1^{\sqrt{3}}\frac{1}{\sqrt t \,\sqrt{1+t^2}}\,dt=\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}-\frac{2 \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{1}{3}\right)}{\sqrt[4]{3}}\approx 0.379133$$ which is still a nightmare ( the Inverse Symbolic Calculator did not find any solution for this number).
Totally unable to find any closed form derivative, I came back to the original problem and expanded the integrand as a Taylor series built at $x=\frac \pi 4$. This gave $$\frac{\sqrt{\tan x}}{\sin x}=\sqrt{2}+\sqrt{2} \left(x-\frac{\pi }{4}\right)^2+\frac{7 \left(x-\frac{\pi }{4}\right)^4}{3 \sqrt{2}}+\frac{139 \left(x-\frac{\pi }{4}\right)^6}{45 \sqrt{2}}+\frac{5473 \left(x-\frac{\pi }{4}\right)^8}{1260 \sqrt{2}}+\cdots$$ Integrating between the given bounds leads to $$I\approx \frac{\pi }{6 \sqrt{2}}+\frac{\pi ^3}{2592 \sqrt{2}}+\frac{7 \pi ^5}{3732480 \sqrt{2}}+\frac{139 \pi ^7}{11287019520 \sqrt{2}}+\frac{5473 \pi ^9}{58511909191680 \sqrt{2}}+\cdots$$ which gives the correct value for six significant figures.
Edit
For sure, the exact solution exits but it involves elliptic integrals. Using the tangent half-angle substitution $t=\tan(\frac x2)$ and the fiven bounds, the result is $$I=-2 i \sqrt{2} \left(F\left(\left.\sin ^{-1}\left(\sqrt[4]{3}\right)\right|-1\right)-F\left(\left.\sin ^{-1}\left(\sqrt{1+\sqrt{2}}\right)\right|-1\right)\right)$$ where appear elliptic integrals of the first kind.
Edit
If we change variable $x=y+\frac \pi 4$, expanding the trigonometric functions and simplifying, we can arrive at $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\sqrt{2}\int_{0}^{\pi /12}\frac{dy}{\sqrt{\cos (2 y)}}=\sqrt{2}\, F\left(\left.\frac{\pi }{12}\right|2\right)$$
$$
\begin{align}
\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\frac{\mathrm{d}\tan(x)}{\sec^2(x)}\tag{1}\\
&=\int_1^\infty\frac{\sqrt{u}}{\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}\tag{2}\\
&=\frac12\int_0^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\tag{3}\\
&=\frac14\int_0^\infty\frac{t^{-3/4}}{(1+t)^{1/2}}\,\mathrm{d}t\tag{4}\\[6pt]
&=\tfrac14\mathrm{B}\left(\tfrac14,\tfrac14\right)\tag{5}\\[6pt]
&=\frac{\frac14\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\tag{6}\\
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)}{\sqrt\pi}\tag{7}\\
\end{align}
$$
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.
We have $$\int_{\pi/4}^{\pi/3}\frac{\sqrt{\tan\left(x\right)}}{\sin\left(x\right)}dx=\int_{\pi/4}^{\pi/3}\frac{1}{\sqrt{\cos\left(x\right)\sin\left(x\right)}}dx=\sqrt{2}\int_{\pi/4}^{\pi/3}\frac{1}{\sqrt{\sin\left(2x\right)}}dx$$ $$=\frac{1}{\sqrt{2}}\int_{\pi/2}^{2\pi/3}\frac{1}{\sqrt{\sin\left(t\right)}}dt $$ and if we put $$\sin\left(t\right)=\cos\left(2u\right) $$ we get $$=\sqrt{2}\int_{0}^{\pi/12}\frac{1}{\sqrt{\cos\left(2u\right)}}dt=\sqrt{2}\int_{0}^{\left(\sqrt{3}-1\right)/2\sqrt{2}}\frac{1}{\sqrt{1-2v^{2}}\sqrt{1-v^{2}}}dt=\sqrt{2}F\left(\frac{\pi}{12}\mid2\right) $$ where $F\left(x\mid k^{2}\right) $ is the elliptic integral of the first kind.
