Solution:

Let 1 (a) be the given integral. By the rule of differentiation under the integral sign.

$\frac{d I}{d a}=\int_{0}^{\pi} \frac{d f}{d a} d x=\int_{0}^{\pi} \frac{1}{\cos x} \cdot \frac{\cos x}{1+\operatorname{acos} x} d x=\int_{0}^{\pi} \frac{d x}{1+\operatorname{acos} x}$

Put $t=\tan \frac{x}{2}, d x=\frac{2 d t}{1+t^{2}}$ and $\cos x=\frac{1-t^{2}}{1+t^{2}}$

When $x=0, t=0$

When $x=\pi, t=\tan \frac{\pi}{2}=\infty$

$\begin{aligned} \therefore \frac{d I}{d a} &=\int_{0}^{\infty} \frac{1}{1+a \cdot\left(\frac{1-t^{2}}{1+t^{2}}\right)} \cdot \frac{2 d t}{1+t^{2}} \\ \frac{d I}{d a} &=\int_{0}^{\infty} \frac{2 d t}{\left(1+t^{2}\right)+a\left(1-t^{2}\right)} \\ \frac{d I}{d a} &=\int_{0}^{\infty} \frac{2 d t}{(1+a)+(1-a) t^{2}} \\ \frac{d I}{d a} &=\frac{1}{1-a} \int_{0}^{\infty} \frac{2 d t}{\left[\frac{1+a}{1-a}\right]+t^{2}} \\ \frac{d I}{d a} &=\frac{2}{1-a} \sqrt{\frac{1-a}{1+a}} \cdot\left[\tan ^{-1} \sqrt{\frac{1-a}{1+a}}\right]_{0}^{\infty} \end{aligned}$

$\begin{aligned} \frac{d I}{d a} &=\frac{2}{\sqrt{1-a^{2}}} \cdot \frac{\pi}{2} \\ \frac{d I}{d a} &=\frac{\pi}{\sqrt{1-a^{2}}} \end{aligned}$

Integrating both sides w.r.t. a, we get

$I=\pi \sin ^{-1} a+c$

To find c, put $a=0$

$I(0)=\pi \sin ^{-1} 0+c, c=0$

$\therefore I=\pi \sin ^{-1} a$

$\therefore \int_{0}^{\pi} \frac{log (1+\operatorname{acos} x)}{cos x} d x=\pi \sin ^{-1} a$