Problem source

Find 
\[
\int_{0}^{\theta}\frac{1}{1-a\cos x}\,\mathrm{d}x\,,
\]
where $0 < \theta < \pi$ and $-1 < a < 1.$
Hence show that 
\[
\int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x=\frac{2}{\sqrt{4-a^{2}}}\tan^{-1}\sqrt{\frac{2+a}{2-a}}\,,
\]
and also that 
\[
\int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x=\frac{\pi}{2}\,.
\]

Solution source

Let $t = \tan \tfrac{x}{2}$, then $\cos x = \frac{1-t^2}{1+t^2}, \frac{d t}{d x} =\tfrac12 (1+t^2)$ so the integral is:

\begin{align*}
\int_0^{\theta} \frac{1}{1-a \cos x} \d x &= \int_{0}^{\tan \frac{\theta}{2}} \frac{1}{1-a \left (\frac{1-t^2}{1+t^2} \right)} \frac{2}{1+t^2} \d t \\
&= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1+t^2-a+at^2} \d t \\
&= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1-a+(1+a) t^2} \d t \\
&= \frac{2}{1+a}\int_0^{\tan \tfrac{\theta}{2}} \frac{1}{\left (\frac{1-a}{1+a} \right)+t^2} \d t \\
&= \frac{2}{1+a} \sqrt{\frac{1+a}{1-a}} \tan^{-1} \left (  \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C \\
&= \frac{2}{\sqrt{1-a^2}}\tan^{-1} \left (  \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C
\end{align*}

Therefore

\begin{align*}
\int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x &= \frac12 \int_0^{\frac12 \pi} \frac{1}{1-\tfrac{a}{2} \cos x} \d x \\
&= \left [\frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\theta}{2} \right) \right]_0^{\pi/2} \\
&= \frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\pi}{4} \right) \\
&= \frac{2}{\sqrt{4-a^2}} \tan^{-1} \left ( \sqrt{\frac{2+a}{2-a} }  \right) \\
\end{align*}

\begin{align*}
\int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= \frac{1}{\sqrt{2}} \int_0^{\frac34 \pi} \frac{1}{1 -\left(- \frac{1}{\sqrt{2}} \right)\cos x} \d x \\
&= \frac{1}{\sqrt{2}} \left [ \frac{2}{\sqrt{1-\tfrac12}} \tan^{-1} \left ( \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} } \tan\frac{\theta}{2} \right) \right]_0^{3\pi/4} \\
&= \frac{1}{\sqrt{2}} \frac{2}{\sqrt{1/2}} \tan^{-1} \left ( \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} } \tan\frac{3\pi}{8} \right) \\
&= 2 \tan^{-1} \left ( \sqrt{\frac{(\sqrt{2}-1)^2}{2-1} } \tan\frac{3\pi}{8} \right)\\
&= 2 \tan^{-1} \left ( (\sqrt{2}-1) \tan\frac{3\pi}{8} \right)
\end{align*}

If $t = \tan \tfrac{3\pi}{8}$, then $-1 = \tan \tfrac{3\pi}{4} = \frac{2t}{1-t^2} \Rightarrow t^2-2t-1 = 0 \Rightarrow t = 1\pm \sqrt{2}$, since $ t > 0$, we must have $t = 1 + \sqrt{2}$, so

\begin{align*}
\int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= 2 \tan^{-1} \left ((\sqrt{2}-1)(\sqrt{2}+1) \right) \\
&= 2 \tan^{-1} 1 \\
&= 2 \frac{\pi}{4} \\
&= \frac{\pi}{2}
\end{align*}