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Evaluate: $$\int_0^{\frac{\pi}{2}} \frac{\arctan{\left(\frac{2\sin{x}}{2\cos{x}-1}\right)}\sin{\left(\frac{x}{2}\right)}}{\sqrt{\cos{x}}} \, \mathrm{d}x$$

I believe there is a "nice" closed form solution but Wolfram is too weak. These arctan integrals are so tricky! I sense a substitution like $\sin{\frac{x}{2}}$ because of arctan argument and $\sqrt{\cos{x}}$ but I just cant get it. Any ideas or tips please.

Source: https://tieba.baidu.com/p/4794735082 (Exercise 3.1.22).

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    $\begingroup$ Why do you believe there is a "nice" closed-form solution for this integral? $\endgroup$
    – an4s
    Aug 3, 2020 at 14:28
  • $\begingroup$ @an4s It's from a website and almost all other integral there have a "nice" closed form. It might not be elementary like it could involve dilogarithm, zeta function, maybe hypergeometric $\endgroup$
    – user801111
    Aug 3, 2020 at 14:36
  • $\begingroup$ Care to share this website that you're referring to? $\endgroup$
    – GohP.iHan
    Aug 3, 2020 at 15:09
  • $\begingroup$ Are you sure that the denominator of the argument of the arctangent function is correct? $\endgroup$
    – Mark Viola
    Aug 3, 2020 at 15:18
  • $\begingroup$ @GohP.iHan tieba.baidu.com/p/4794735082 $\endgroup$
    – user801111
    Aug 3, 2020 at 16:20

1 Answer 1

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$$\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx \overset{\tan\frac{x}{2}\to x}= 2\int_0^1 \frac{x\arctan\left(\frac{4x}{1-3x^2}\right)}{\sqrt{1-x^2}(1+x^2)}dx$$ $$=2\underbrace{\int_0^1 \frac{x(\arctan x + \arctan(3x))}{\sqrt{1-x^2}(1+x^2)}dx}_{\mathcal J} - 2\pi \underbrace{\int_\frac{1}{\sqrt 3}^1 \frac{x}{\sqrt{1-x^2}(1+x^2)}dx}_{\large\frac{1}{2\sqrt2}\ln(2+\sqrt 3)} $$ The second integral appears since for $x>\frac{1}{\sqrt 3}$ we have $x \cdot 3x>1$.
Now in order to deal with $\mathcal J$ we can differentiate under the integral sign, considering: $$\mathcal J(a)=\int_0^1\frac{x(\arctan(a x)+\arctan(3 x))}{\sqrt{1-x^2}(1+x)}dx $$ $$\Rightarrow \mathcal J'(a) = \int_0^1 \frac{x^2}{\sqrt{1-x^2}(1+x^2)(1+a^2 x^2)}dx \overset{\frac{1}{x}\to \sqrt{1+x^2}}=\int_0^\infty \frac{dx}{(2+x^2)(1+a^2+x^2)}$$ $$=\frac{2}{1-a^2}\int_0^\infty \left(\frac{1}{1+a^2+x^2}-\frac{1}{2+x^2}\right)dx=\frac{\pi}{2(1-a^2)}\left(\frac{1}{\sqrt{1+a^2}}-\frac{1}{\sqrt 2}\right)$$ We are looking to find $\mathcal J(1)=\mathcal J$, but $\mathcal J(-3)=0$ therefore our integral is: $$\mathcal J=\frac{\pi}{2}\int_{-3}^1 \frac{1}{1-a^2}\left(\frac{1}{\sqrt{1+a^2}}-\frac{1}{\sqrt 2}\right)da\overset{a=\frac{1+x}{1-x}}=\frac{\pi}{4\sqrt 2}\int_0^2\frac{1}{x}\left(1-\frac{1-x}{\sqrt{1+x^2}}\right)dx$$ $$=\frac{\pi}{4\sqrt 2}\left(\ln\left(x+\sqrt{1+x^2}\right)+\ln\left(1+\sqrt{1+x^2}\right)\right)\bigg|_0^2=\frac{\pi}{\sqrt 2} \ln \varphi,\ \varphi =\frac{1+\sqrt 5}{2}$$ Therefore: $$\boxed{\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx=\sqrt 2 \pi \ln\varphi-\frac{\pi}{\sqrt 2}\ln(2+\sqrt 3)}$$

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