Hint for solving a definite integral I stumbled upon this problem which I cant seem to solve : 
Prove that : 
$$\int_0^\frac{\pi}{2}\frac{\log(1+\cos\alpha \cos x)}{\cos x}dx=\frac{1}{2}\left[\frac{\pi^2}{4}-\alpha^2\right]$$
I've tried all that i could.
Any solution for this?
If not a solution, maybe a hint or a direction in which i could think?   

Let 
$$I (\alpha) = \int_0^{\pi/2} \frac{\ln (1 + \cos \alpha \cos x)}{\cos x} \, dx, \quad \alpha \in \mathbb{R}.$$
Note that $I(\frac{\pi}{2}) = 0$. Now differentiating with respect to the parameter $\alpha$ we have
$$I'(\alpha) = - \sin \alpha \int_0^{\pi/2} \frac{dx}{1 + \cos \alpha \cos x}.$$
The resulting integral can be evaluated using a Weierstrass substitution of $t = \tan (x/2)$. Doing so yields
$$I'(\alpha) = - 2\sin \alpha \int_0^1 \frac{dt}{1 + \cos \alpha + (1 - \cos \alpha) t^2}.$$
Now as the term $1 - \cos \alpha \geqslant 0$ for all $\alpha \in \mathbb{R}$ we can write
\begin{align*}
I'(\alpha) &= -\frac{2 \sin \alpha}{1 - \cos \alpha} \int_0^1 \frac{dt}{ \left (\sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} \right )^2 + t^2}\\
&= -\frac{2 \sin \alpha}{1 - \cos \alpha} \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} \tan^{-1} \left (\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} \right )\\
&= -\alpha,
\end{align*}
where in the last line the following well-known result for the inverse tangent function has been used
$$\cos^{-1} (x) = 2 \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ), \quad -1 < x \leqslant 1.$$
So on integrating up with respect to $\alpha$
we have
$$I(\alpha) = -\frac{\alpha^2}{2} + C,$$
where $C$ is an arbitrary constant to be determined. Since $I(\frac{\pi}{2}) = 0$, one finds $C = \pi^2/8$. Thus
$$I(\alpha) = \frac{1}{2} \left (\frac{\pi^2}{4} - \alpha^2 \right ),$$
as required. 
