How to evaluate or prove this integral:$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt[4]{2\sec^2x+\sqrt3}}dx=\frac{\pi}{3\sqrt[8]{3}}$?

Question

This is my question:

How to evaluate or prove this integral: \begin{equation} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt[4]{2\sec^2x+\sqrt3}}dx=\frac{\pi}{3\sqrt[8]{3}} \end{equation}

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$\textbf{My Attempt:}$
Let : $\displaystyle I =\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt[4]{2\sec^2x+\sqrt3}}dx$, and we have:

\begin{aligned} I &=\frac{1}{\sqrt[8]{3}}\int_0^{\frac{\pi}{2}}\frac{\sqrt {\cos x}}{\sqrt[4]{\frac{2}{\sqrt3}+\cos ^2x}}dx \\ &=\frac{1}{\sqrt[8]{3}}\int_0^{\frac{\pi}{2}}\frac{{\cos x}}{\sqrt[4]{\frac{2}{\sqrt3}\cos ^2x+\cos ^4x}}dx\\ &=\frac{1}{\sqrt[8]{3}}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt[4]{\frac{2}{\sqrt3}(1-\sin^2x)+(1-\sin ^2x)^2}}d(\sin x)\\ &=\frac{1}{\sqrt[8]{3}}\int_0^{1}\frac{1}{\sqrt[4]{\frac{2}{\sqrt3}(1-x^2)+(1-x^2)^2}}dx \end{aligned}

At this point, I couldn't get any more efficient calculations, maybe we'll use elliptic integrals?

If I can prove

$$\displaystyle\int_0^{1}\frac{1}{\sqrt[4]{\frac{2}{\sqrt3}(1-x^2)+(1-x^2)^2}}dx=\frac{\pi}{3}$$ it's the best.

Could anybody give me some hints? Thanks!

Answer 1

A Short Comment(it's better to have a general form of the hypergeometric function in terms of elliptic integrals.)
Recall this integral representation of Gauss hypergeometric series, $$ \int_{0}^{1} x^{b-1}(1-x)^{c-b-1}\left ( 1-tx \right )^{-a} \text{d}x =\frac{\Gamma\left ( b \right )\Gamma\left ( c-b \right ) }{ \Gamma\left (c \right ) } \,_2F_1\left ( a,b;c;t \right ). $$ Now that you have the last integral formulation, one can reduce it into a convenient form, $$ I=\int_{0}^{1} \frac{1}{\sqrt[4]{1-x^2} \sqrt[4]{\left ( 2+\sqrt{3} \right )/\sqrt{3} -x^2 } } \text{d}x =\frac{\sqrt{2}\sqrt[4]{\sqrt{3}\left ( 2-\sqrt{3} \right ) } }{\sqrt{\pi} } \Gamma\left ( \frac34 \right )^2 \,_2F_1\left ( \frac14,\frac12;\frac54;\sqrt{3} \left ( 2-\sqrt{3} \right ) \right ). $$ Note that the above expression could be written as follows, $$ I=\frac{\sqrt{2}\,\Gamma\left ( \frac34 \right )^2 }{\sqrt{\pi} } \int_{0}^{\sqrt[4]{\sqrt{3}\left ( 2-\sqrt{3} \right )}} \frac{1}{\sqrt{1-x^4} }\text{d}x. $$ Make the substitution $x=\frac{1}{\sqrt{z}}$, then $$ \int_{0}^{\alpha} \frac{1}{\sqrt{1-x^4} }\text{d}x =\frac12\int_{\frac1{\alpha^2}}^{\infty} \frac{1}{\sqrt{x^3-x} } \text{d}x,\qquad \alpha=\sqrt[4]{\sqrt{3}\left ( 2-\sqrt{3} \right )}. $$ Employing the Weierstrass elliptic function $\wp(z)$ with $g_2=1,g_3=0$ and $4\wp(z)^3-g_2\wp(z)-g_3=\wp^\prime(z)^2$, we write $$ I=\frac{\Gamma\left ( \frac34 \right )^2 }{\sqrt{\pi} }\int_{\frac12\sqrt{\frac{2+\sqrt{3} }{\sqrt{3}} } }^{\infty} \frac{1}{\sqrt{4x^3-x} } \text{d}x=\frac{\Gamma\left ( \frac34 \right )^2 }{\sqrt{\pi} }\omega, $$ where $\wp(\omega)=\frac12\sqrt{\frac{2+\sqrt{3} }{\sqrt{3}} }$. The real half-period is related to one of the roots $4x^3-x=0\rightarrow x=\frac12$, given by $\frac{\omega_0}{2} =\int_{\frac12}^{\infty} \frac{1}{\sqrt{4x^3-x} } \text{d}x= \frac{\Gamma\left ( \frac14 \right )^2}{4\sqrt{\pi} }$. One may use the addition formula to evaluate the 3-torsion point $\wp(\omega_0/3)=\frac12\sqrt{\frac{2+\sqrt{3} }{\sqrt{3}} }$, where $z_1,z_2$ are arbitrary complex numbers that make sense, $$\wp\left(z_1-z_2;g_2,g_3\right)\wp\left(z_1+z_2;g_2,g_3\right)=\frac{\left(\wp\left(z_1;g_2,g_3\right)\wp\left(z_2;g_2,g_3\right)+\frac{g_2}{4}\right){}^2+g_3\left(\wp\left(z_1;g_2,g_3\right)+\wp\left(z_2;g_2,g_3\right)\right)}{\left(\wp\left(z_1;g_2,g_3\right)-\wp\left(z_2;g_2,g_3\right)\right){}^2}.$$ In our case it is $$ \wp\left ( z_1+z_2 \right ) \wp\left ( z_1-z_2 \right ) =\left ( \frac{\wp\left ( z_{1} \right )\wp\left ( z_{2} \right ) +\frac14}{\wp\left ( z_{1} \right )-\wp\left ( z_{2} \right )} \right )^2. $$ While $\wp(\omega_0/2)=\frac12$ has been evaluated, we substitute $(z_1,z_2)=(\omega_0/3,\omega_0/6)$ and $(z_1,z_2)=(\omega_0/2,\omega_0/3)$ respectively, giving $$ \wp\left(\frac{\omega_0}{2} \right ) \wp\left ( \frac{\omega_0}{6}\right ) =\left ( \frac{\wp\left ( \frac{\omega_0}{3}\right )\wp\left ( \frac{\omega_0}{6} \right ) +\frac14}{\wp\left ( \frac{\omega_0}{3} \right )-\wp\left (\frac{\omega_0}{6} \right )} \right )^2, $$ $$ \wp\left(\frac{\omega_0}{6} \right ) \wp\left ( \frac{5\omega_0}{6}\right ) =\left ( \frac{\wp\left ( \frac{\omega_0}{2}\right )\wp\left ( \frac{\omega_0}{3} \right ) +\frac14}{\wp\left ( \frac{\omega_0}{2} \right )-\wp\left (\frac{\omega_0}{3} \right )} \right )^2. $$ By periodicity, $\wp\left ( \frac{5\omega_0}{6}\right )=\wp\left ( \frac{-\omega_0}{6}\right )$, and it's an even function, we have $\wp\left ( \frac{5\omega_0}{6}\right )=\wp\left ( \frac{\omega_0}{6}\right )$. Solving the system we obtain $$ \wp\left ( \frac{\omega_0}{3}\right ) =\frac{1}{2} \sqrt{\frac{2+\sqrt{3} }{\sqrt{3} } } ,\wp\left ( \frac{\omega_0}{6}\right ) =\frac{1}{2} \left ( 1+\sqrt{3}+\sqrt{3+2\sqrt{3} } \right ). $$ Hence we have $$ I=\frac{\Gamma\left ( \frac34 \right )^2 }{\sqrt{\pi} } \cdot\frac{1}{3} \frac{\Gamma\left ( \frac14 \right )^2}{2\sqrt{\pi} } =\frac{\pi}{3} . $$

Answer 2

Partial solution

Let the integral be $I$. \begin{align*} I &= \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt[4]{2 \sec^2 x + \sqrt{3}}} dx \\ &= \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt[4]{\frac{2}{\cos^2 x} + \sqrt{3}}} dx \\ &= \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt[4]{2 + \sqrt{3} \cos^2 x}} dx \end{align*}

\begin{align*} I &= \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sqrt[4]{(2 + \sqrt{3} \cos^2 x) \cos^2 x \sqrt{\sin^2 x}}} dx \\ &= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{\sqrt[4]{(2 + \sqrt{3}(1-\sin^2 x))(1-\sin^2 x) \sqrt{\sin^2 x}}} dx \end{align*} Let $y = \sin^2 x$. For $x \in (0, \frac{\pi}{2})$, we have $dy = 2 \sin x \cos x dx$. The limits of integration change accordingly: when $x=0$, $y=0$; when $x=\frac{\pi}{2}$, $y=1$. Substituting these into the integral: \begin{align*} I &= \frac{1}{2} \int_{0}^{1} \frac{dy}{\sqrt[4]{(2+\sqrt{3}(1-y))(1-y)\sqrt{y}}} \\ &= \frac{1}{2} \int_{0}^{1} \frac{dy}{\sqrt[4]{((2+\sqrt{3}) - \sqrt{3}y)(1-y)y^{\frac{1}{2}}}} \\ &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \int_{0}^{1} \left(1 - \frac{\sqrt{3}y}{2+\sqrt{3}}\right)^{-\frac{1}{4}} (1-y)^{-\frac{1}{4}} y^{-\frac{1}{8}} dy \end{align*} Now, expand the term $\left(1 - \frac{\sqrt{3}y}{2+\sqrt{3}}\right)^{-\frac{1}{4}}$ using the generalized binomial theorem $(1-z)^{-\alpha} = \sum_{n=0}^{\infty} \frac{(\alpha)_n}{n!} z^n$, where $(\alpha)_n = \alpha(\alpha+1)\dots(\alpha+n-1)$. The series expansion is: $$ \left(1 - \frac{\sqrt{3}y}{2+\sqrt{3}}\right)^{-\frac{1}{4}} = \sum_{n=0}^{\infty} \frac{1}{n!} \left(-\frac{1}{4}\right)\left(-\frac{1}{4}-1\right)\dots\left(-\frac{1}{4}-n+1\right) \left(-\frac{\sqrt{3}y}{2+\sqrt{3}}\right)^n $$ This simplifies to $\sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{n! 4^n} \left(\frac{\sqrt{3}y}{2+\sqrt{3}}\right)^n$. Substituting this back into the integral expression for $I$: \begin{align*} I &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{n! 4^n} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \int_{0}^{1} y^n (1-y)^{-\frac{1}{4}} y^{-\frac{1}{8}} dy \\ &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{n! 4^n} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \int_{0}^{1} y^{n-\frac{1}{8}} (1-y)^{-\frac{1}{4}} dy \end{align*} The integral $\int_{0}^{1} y^{n-\frac{1}{8}} (1-y)^{-\frac{1}{4}} dy$ is the Beta function $B\left(n-\frac{1}{8}+1, -\frac{1}{4}+1\right) = B\left(n+\frac{7}{8}, \frac{3}{4}\right)$.

\begin{align*} I &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{n! 4^n} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n B\left(n+\frac{1}{8}, \frac{3}{4}\right) \end{align*} Expressing the Beta function using Gamma functions, $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. \begin{align*} I &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{n! 4^n} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \frac{\Gamma(n+\frac{1}{8})\Gamma(\frac{3}{4})}{\Gamma(n+\frac{5}{4})} \end{align*} Using the identity $\frac{1 \cdot 5 \cdot 9 \dots (4n-3)}{4^n} = \frac{\Gamma(n+\frac{1}{4})}{\Gamma(\frac{1}{4})}$: \begin{align*} I &= \frac{1}{2\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{4})}{n!\Gamma(\frac{1}{4})} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \frac{\Gamma(n+\frac{1}{8})\Gamma(\frac{3}{4})}{\Gamma(n+\frac{5}{4})} \\ &= \frac{\Gamma(\frac{3}{4})}{2\Gamma(\frac{1}{4})\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{\Gamma(n+\frac{1}{4})\Gamma(n+\frac{1}{8})}{\Gamma(n+\frac{5}{4})} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \end{align*}

\begin{align*} I &= \frac{2\sqrt{\pi}}{\sqrt[4]{2+\sqrt{3}}} \frac{\Gamma^2(\frac{3}{4})}{\Gamma(\frac{1}{4})\sin\frac{\pi}{4}} \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(4n+1)2^n n!} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \\ &= \frac{\sqrt{2}\Gamma^2(\frac{3}{4})}{\sqrt{\pi}\sqrt[4]{2+\sqrt{3}}} \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(4n+1)2^n n!} \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^n \end{align*}

I don’t know how to proceed further.