Integral$ \int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$ $$\int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$$
I tried to substitute $1+\cos^2x$ , tried to change cos in sin by complementary formula etc . But nothing seems to work out. I think it's not as simple as I thought ?

$$ I = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1+\cos^2 x}} dx = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{2-\sin^2 x}} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx $$
Recall the elliptic integral of the first and second kind, respectively:
$$F(\phi,k) = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}}d\theta$$
$$E(\phi,k) = \int_{0}^{\phi} \sqrt{1-k^2\sin^2 \theta}d\theta$$
Is easy to show that
$$\int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{F(\phi,k)-E(\phi,k)}{k^2}$$
$$\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{1}{k^2}\int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2\phi}} - \sqrt{1-k^2\sin^2\phi} d\phi=  \frac{1}{k^2}\int_{0}^{\phi} \left[\frac{1}{\sqrt{1-k^2\sin^2\phi}} -\frac{1-k^2\sin^2 \phi}{\sqrt{1-k^2\sin^2\phi}}\right]d\phi =  \int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta$$
Hence:
$$ \int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1-\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta - \int_{0}^{\phi} \frac{\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta  = F(\phi,k) -\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{E(\phi,k)-(1-k^2)F(\phi,k)}{k^2} = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}$$
where $k$ is the modulus and $k'$ is the complementary modulus and $|k|<1, \quad k=\sqrt{1-k'^2}$.
Hence
$$\boxed{\int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}}$$
Therefore
$$\boxed{I = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx = \frac{2}{\sqrt{2}}\left[E\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)-\frac{1}{2}F\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)\right]}$$
We have to take care evaluating elliptic functions since the software sometimes may use a slight different notation
