Integration

$I = \large \int_0^\pi \dfrac{ \ln (1 + \cos x)} { \cos x} \, dx$ $\Rightarrow I = \large \int_0^\frac{\pi}{2} \dfrac{ \ln (1 + \cos x)} { \cos x} \, dx - \large \int_0^\frac{\pi}{2} \dfrac{ \ln (1 - \cos x)} { \cos x} \, dx$ $\Rightarrow I = \large \int_0^\frac{\pi}{2} \dfrac{ \ln (\frac{1 + \cos x}{1- \cos x})} { \cos x} \, dx$ $\Rightarrow I = -\large \int_0^\frac{\pi}{2} \dfrac{\ln (\tan^2\frac{x}{2})}{ \cos x} \, dx$ Put $\tan\frac{x}{2} = t$ $\Rightarrow I = -4\large \int_0^1 \dfrac{\ln t}{1 - t^2} \, dt$ $\Rightarrow I = -4\large \int_0^1 \ln t \sum_{k=0}^{\infty} t^{2k} \, dt$ Switching order of summation and integration, $\Rightarrow I = -4\sum_{k=0}^{\infty}\large \int_0^1 t^{2k}\ln t \, dt$ Now $\large \int_0^1 t^{2k}\ln t \, dt = -\frac{1}{(2k+1)^2}$ $\Rightarrow I = 4\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}$ Since $\sum_{k=0}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$ , $\sum_{k=0}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4}\sum_{k=0}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{24}$ and $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4}\sum_{k=0}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{8}$ $\Rightarrow I = 4\frac{\pi^2}{8} = \frac{\pi^2}{2}$

$\begin{aligned} &I(a)=\int_0^\pi \frac{\ln(1+a\cdot cos(x))}{\cos(x)}\, dx\\ &{I}'(a)=\int_0^\pi \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{1+a\cdot cos(x)}\, dx && {\color{#D61F06} \text{differentiate both sides with respect to a}}\\ &{I}'(a)=\int_0^\pi \frac{1}{1+a\cdot cos(x)}\, dx \\ &{I}'(a)=\int_0^\infty \frac{1}{1+a\frac{1-t^2}{1+t^2}}\cdot \frac{2dt}{1+t^2}&&{\color{#D61F06} t=tan(\frac{x}{2}) \, , \cos(x)=\frac{1-t^2}{1+t^2} \, , dx=\frac{2dt}{1+t^2} , \text{weierstrass substitution}}\\ &{I}'(a)=2\int_0^\infty \frac{dt}{(1+a)+(1-a)t^2}\\ &{I}'(a)=\frac{2}{1+a}\int_0^\infty \frac{dt}{1+\left ( \sqrt{\frac{1-a}{1+a}} \cdot t \right )^2}\\ &{I}'(a)=\frac{2}{\sqrt{1-a^2}}\int_0^\infty \frac{du}{1+u^2} && {\color{#D61F06} u=\sqrt{\frac{1-a}{1+a}} \cdot t, dt=\sqrt{\frac{1+a}{1-a}}\cdot du}\\ &{I}'(a)=\frac{2}{\sqrt{1-a^2}} \left . \tan^{-1}(u) \right |_0^\infty=\frac{\pi}{\sqrt{1-a^2}}\\ &I(a)=\pi \sin^{-1}(a)+C && {\color{#D61F06} \text{integrate both sides}}\\ &\color{#D61F06} a=0 \Rightarrow I(0)=\int_0^\pi \frac{\ln(1)}{\cos(x)}\, dx =0 \\ &\color{#D61F06} I(0)=\pi \sin^{-1}(0)+C = 0 \Rightarrow C=0 \\ &I(a)=\pi \sin^{-1}(a) \\ &I(1)=\int_0^\pi \frac{\ln(1+cos(x))}{\cos(x)}\, dx=\pi \sin^{-1}(1)\\ & \int_0^\pi \frac{\ln(1+cos(x))}{\cos(x)}\, dx= \frac{\pi^2}{2}\approx 4.9348 \end{aligned}$