Prove via differentiation that integral $\int_0^\pi \frac{\log(1+\cos\alpha\cos\theta)}{\cos\theta}\,d\theta = \pi(\frac{\pi}{2}-\alpha) $ 
Prove via differentiation that integral 
  $$\int_0^\pi \frac{\log(1+\cos\alpha\cos\theta)}{\cos\theta}\,d\theta$$
  is equal to
  $$ \pi\left(\frac{\pi}{2} - \alpha\right) $$
  where $0\leq \alpha\leq \frac{\pi}{2}.$

After partially differentiating wrt $\alpha$ I have tried via substitution;
$$u=\log(1+\cos\alpha\cos\theta)$$
Then I get;
$$\frac{dI(\alpha)}{d\alpha}=-\sin\alpha\int_0^\pi \frac{1}{1+\cos\alpha\cos\theta}\,d\theta$$
I can not currently see how to progress this to stated answer.
Please show full working.
The question is from G Stephenson's 'Mathematical Methods for Science students' - page 172.
 A: Let $I(\alpha)$ be the integral of interest given by
$$I(\alpha)=\int_0^\pi \frac{\log(1+\cos(\alpha)\cos(\theta))}{\cos(\theta)}\,d\theta$$
Differentiating $I(\alpha)$, we obtain
$$\frac{dI(\alpha)}{d\alpha}=-\sin(\alpha)\int_0^\pi\frac{1}{1+\cos(\alpha)\cos(\theta)}\,d\theta$$
Note that we can use the classical Tangent Half-Angle Substitution to evaluate the integral 
$$f(\alpha)=\int_0^\pi \frac{1}{1+\cos(\alpha)\cos(\theta)}\,d\theta=\frac{\pi}{\sin(\alpha)}$$
Therefore, we have $\frac{dI(\alpha)}{d\alpha}=-\pi$ whereupon integrating yields
$$I(\alpha)=-\pi \alpha +C$$
Noting that $I(\pi/2)=0$ reveals
$$I(\alpha)=\pi(\pi/2-\alpha)$$
as was to be shown!

NOTE:
To arrive at the closed-form expression for $f(\alpha)$ we enforce the substitution $t=\tan(x/2)$.  Then, $\cos(x)=\frac{1-t^2}{1+t^2}$, $dx=\frac{2}{1+t^2}\,dt$, and the limits extend from $t=0$ to $t=\infty$ to reveal
$$\begin{align}
f(\alpha)&=\frac{1}{\sin^2(\alpha/2)}\int_0^\infty \frac{1}{\cot^2(\alpha/2)+t^2}\,dt\\\\
&=\frac{1}{\sin^2(\alpha/2)}\frac{1}{\cot(\alpha/2)}\left.\left(\arctan\left(\frac{t}{\cot(\alpha/2)}\right)\right)\right|_{0}^{\infty}\\\\
&=\frac{1}{\frac12\sin(\alpha)}  \,\frac{\pi}{2}\\\\
&=\frac{\pi}{\sin(\alpha)}
\end{align}$$ 
