MHB Evaluation of definite Integral

[juantheron]

$\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi}{2}-\int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx$Using The formula $\displaystyle \tan^{-1}(x)+\cot^{-1}(x) = \frac{\pi}{2}\Rightarrow \tan^{-1}(x) = \frac{\pi}{2}-\cot^{-1}(x).$Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{\cos^2 x-\sin^2 x}{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\cot^{-1}\sqrt{\frac{1}{2}-\frac{\tan^2 x}{2}}dx$Now How can i solve after thatThanks

 

[Saitama]

Did anyone solve this?

I found the same problem on stackexchange and it has been unanswered from four days.

Wolfram Alpha gives 0.411234 which is quite close to $\zeta(2)/4$. I hope somebody can confirm this by using a better software.

Looking at the answer, I feel it requires the use of series expansion. $\arctan(x)$ has got a nice series expansion but I am not sure if that would help. I attempted it the following way:
$$I=\int_0^{\pi/4} \arctan\left(\sqrt{\frac{\cos (2x)}{1+\cos(2x)}}\right)\,dx=\frac{1}{2}\int_0^{\pi/2} \arctan\left(\sqrt{\frac{\cos (x)}{1+\cos(x)}}\right)\,dx$$
$$\Rightarrow I=\frac{1}{2}\int_0^{\pi/2} \arctan\left(\sqrt{\frac{\sin (x)}{1+\sin(x)}}\right)\,dx=\frac{1}{4}\int_0^{\pi} \arctan\left(\sqrt{\frac{\sin (x)}{1+\sin(x)}}\right)\,dx$$
But I don't see a way to proceed after this.

I am curious to see a solution to this problem.

@jacks: From where did you get this problem?

 

[alyafey22]

Try the following substitution

tan substitution

Then use integration by parts to kill the inverse tangent.

 

[juantheron]

To pranav I have seen somewhere in another forum.

Here is a solution given by sos440.

? ? ?? ? ?? :: ? ?? 37 - Coxeter's Integrals (1)

 

[Saitama]

To pranav I have seen somewhere in another forum.

Here is a solution given by sos440.

? ? ?? ? ?? :: ? ?? 37 - Coxeter's Integrals (1)

Thanks jacks! Sos solution is great! :)

Oh and the answer is $\zeta(2)/4$ as I said before. :p