Integration with respect to a parameter $a$

Question

Show that the value of the integral of $$\frac{\log(1+\cos(a)\cos(x))}{\cos(x)}$$ over the interval $[0,\pi]$ is given by $$\pi\left(\frac{\pi}{2} - a\right)$$ for $0\leq\,a\leq\,2.$

Answer 1

Let's start by defining $$ I_a = \int_0^\pi\frac{\log(1+\cos(a)\cos(x))}{\cos(x)}\,dx. $$ Observe that $I_{\pi/2} = 0$, so we would be done if we could prove that $I_{a}' = -\pi$, where the derivative is with respect to $a$. So taking the derivative under the integral sign, we get $$ I_a' = -\sin(a) \int_0^\pi \frac{dx}{1+\cos(a)\cos(x)}. $$ Performing the change of variables $y=\pi-x$, we can also see that $$ I_a' = -\sin(a) \int_0^\pi \frac{dx}{1-\cos(a)\cos(x)}. $$ Taking the average of the two integrals, and summing the fractions we arrive at $$ I_a' = -\sin(a) \int_0^\pi \frac{dx}{1-\cos^2(a)\cos^2(x)} = -2\sin(a)\int_0^{\pi/2} \frac{dx}{1-\cos^2(a)\cos^2(x)}. $$ In the second step we used the symmetry of the integrand with respect to the swap $x\mapsto \pi-x$.

To proceed with this integral, we now use the change of coordinates $u=\tan(x)$. Then $\cos^2(x) = (1+u^2)^{-1}$ and $(1+u^2)^{-1}du = dx$, so $$ I_a' = -2\sin(a)\int_0^\infty\frac{du}{u^2+1-\cos^2(a)} = -2\sin(a)\int_0^\infty \frac{du}{u^2+\sin^2(a)}. $$ Now we make another change of variables $u = \sin(a)v$ to simplify further: $$ I_a' = -2\int_0^\infty \frac{dv}{v^2+1} = -\pi. $$ where the last equality can be proven using the trigonometric substitution $v = \tan(\theta)$.