$$\displaystyle\int_0^{\frac{\pi}{4}}{\,x}{\,\arctan\sqrt{\frac{\cos2x}{2\sin^2x}}}dx=\frac{\pi}{96}[{\pi^2}-6\ln^22]$$ And $$\displaystyle\int_0^{\frac{\pi}{4}}{\,x}{\,\arctan\sqrt{\frac{2\sin^2x}{\cos2x}}}dx=\frac{\pi}{192}[{\pi^2}+12\ln^22]$$ These integrals have been proposed by my friend,but I do not know how to proceed. How do you evaluate these integral?

  • 4
    $\begingroup$ Ask your friend...? $\endgroup$
    – Timbuc
    Dec 19, 2014 at 22:52
  • $\begingroup$ $1-\cos2x=2\sin^2x$. $\endgroup$
    – Lucian
    Dec 20, 2014 at 1:21
  • $\begingroup$ @Venus: I really doubt that the linked technique would work for this problem. Instead, the problem at hand can be simplified to $\int_0^{\pi/4} x\,\arcsin\sqrt{\cos(2x)}\,dx$ and I guess it is easier to start with this form but I haven't yet succeeded in solving the problem. $\endgroup$ Dec 20, 2014 at 13:07

1 Answer 1


Denote the first integral by $I$ and the second by $J$. Then,

$$\begin{aligned} J=&\int_0^{\pi/4} x\left(\frac{\pi}{2}-\arctan\sqrt{\frac{\cos 2x}{2\sin^2 x}}\right)\,dx \\ =&\frac{\pi^3}{64}-I \,\,\,\,\,\,\,(1) \end{aligned}$$

$I$ can be simiplified to: $$I=\int_0^{\pi/4} x\arccos(\sqrt{2}\sin x)\,dx=\left(\frac{x^2\arccos(\sqrt{2}\sin x)}{2}\right|_0^{\pi/4}+\frac{1}{2}\int_0^{\pi/4} \frac{x^2\left(\sqrt{2}\cos x\right)}{\sqrt{1-2\sin^2x}}\,dx$$ The first term is zero and with the substitution $2\sin^2x=\sin^2\theta$, $$I=\frac{1}{2}\int_0^{\pi/2} \left(\arcsin\left(\frac{\sin \theta}{\sqrt{2}}\right)\right)^2\,d\theta$$ From here, $$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}x^{2n+1}$$ Integrate both sides within the limit $0$ to $\sin\theta/\sqrt{2}$, i.e: $$\begin{aligned} I &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}\frac{(2n)!!}{(2n+1)!!}\int_0^{\pi/2} \sin^{2n+2}\theta\,d\theta \\ &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}\frac{(2n)!!}{(2n+1)!!}\frac{(2n+1)!! \pi}{2^{n+2}(n+1)!}\\ &=\frac{\pi}{8}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}(n+1)^2} \,\,\,\,\,\,\,\,\,\left((2n)!!=2^nn!\right) \\ &=\frac{\pi}{8}\text{Li}_2\left(\frac{1}{2}\right) \\ \end{aligned}$$



and from $(1)$,

$$\boxed{J=\dfrac{\pi}{192}\left(\pi^2+12\ln^2 2\right)}$$

  • 2
    $\begingroup$ +1. A clean and straightforward answer. I think it might be possible to generalise this integral to $$I(a,b,r,s)=\int^\frac{\pi}{2}_0\frac{\arctan(r\sin{x})\arctan(s\sin{x})}{a^2-2ab\cos(2x)+b^2}{\rm d}x$$ $\endgroup$
    – M.N.C.E.
    Dec 21, 2014 at 7:58
  • $\begingroup$ Very,very nice.Thank you $\endgroup$
    – user178256
    Dec 21, 2014 at 12:50
  • 1
    $\begingroup$ @user178256: Please consider accepting the answer. You can do that by clicking the tick near the start of my answer. $\endgroup$ Dec 21, 2014 at 14:09

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