$$\int_0^{\frac\pi2}\frac{dx}{1+\sin^2(\tan x)}$$

First, I tried to set

$$t=\tan x$$

Then I got $$\int_0^\infty\frac{dt}{(1+t^2)(1+\sin^2t)}$$

applied a trig identity, $1+ \sin^2t=\frac{1}{2}(3-\cos2 t)$

I got

$$\int_0^\infty\frac{dt}{(1+t^2)\left(\frac12(3-\cos 2t)\right)}$$

I don't know how to keep going to solve it.

can some one give me a hint?



marked as duplicate by Simple, Aditya Hase, Venus, user147263, drhab Dec 29 '14 at 9:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ $1+ \sin^2t=\frac{1}{2}(3-\cos t)$? it should be $1+ \sin^2t=\frac{1}{2}(3-\cos 2t)$ $\endgroup$ – Krish Dec 29 '14 at 8:30
  • 1
    $\begingroup$ @Krish yes, you are right. But I still don't have any good idea to solve it $\endgroup$ – Simple Dec 29 '14 at 8:32
  • $\begingroup$ Do you want to have a closed form solution to this integral? Because after looking at this Wolfram alpha answer, I am not sure there is one. $\endgroup$ – Samrat Mukhopadhyay Dec 29 '14 at 8:33
  • 1
    $\begingroup$ @M.N.C.E. thanks for sharing the link $\endgroup$ – Simple Dec 29 '14 at 8:44