Evaluating $\int_{\pi/4}^ { \pi/2} \frac{1}{\sqrt{1+\cos²(x)}} \text{d}x$? I want to evaluate this integral 
$$\int_{\pi/4}^ { \pi/2}  \frac{1}{\sqrt{1+\cos^2(x)}} \text{d}x  $$
but I don't know how to proceed.
 A: Rewrite slightly (using $\cos^2+\sin^2=1$):
$$
I=\frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\frac{1}{\sqrt{1-\frac{1}{2}\sin^2(x)}}dx$$
Looking up wikipedia this is the standardform of $\text{F}(x|a)$, ( elliptic integrals of the first typ) with $a=\sqrt\frac{1}{2}$ .
$$
I=\frac{1}{\sqrt{2}}\text{F}(\frac{1}{\sqrt{2}})-\frac{1}{\sqrt{2}}\text{F}(\frac{\pi}{4}|\frac{1}{\sqrt{2}})
$$
A: By setting $x=\arctan u$ we are left with:
$$ \int_{1}^{+\infty}\frac{du}{(1+u^2)\sqrt{1+\frac{1}{1+u^2}}}=\int_{1}^{+\infty}\frac{du}{\sqrt{(1+u^2)(2+u^2)}}$$
that can be written in terms of elliptic integrals, namely as a difference between a complete elliptic integral of the first kind and an incomplete elliptic integral of the second kind. Unluckily, I do not believe that can be simplified further. As a positive counterpart, the AGM gives a very fast-converging algorithm for the numerical evaluation of such integrals. Ultimately, that boils down to noticing that:
$$ (1+u^2)(2+u^2) = ((\sqrt{2}-1)u)^2 + (u^2+\sqrt{2})^2$$
by Lagrange's identity.
