Problem source

Show that 
\begin{questionparts}
\item $\dfrac{1-\cos\alpha}{\sin\alpha}=\tan\frac{1}{2}\alpha,$ 
\item  if $\left|k\right|<1$ then ${\displaystyle \int\frac{\mathrm{d}x}{1-2kx+x^{2}}=\frac{1}{\sqrt{1-k^{2}}}\tan^{-1}\left(\frac{x-k}{\sqrt{1-k^{2}}}\right)+C,}$
where $C$ is a constant of integration. 
\end{questionparts}
Hence, or otherwise, show that if $0<\alpha<\pi$ then 
\[
\int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x=\frac{\pi-\alpha}{2}.
\]

Solution source

\begin{questionparts}
\item \begin{align*}
\frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\
&= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\
&= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\
&= \tan \tfrac{\alpha}{2}
\end{align*}
\item \begin{align*}
\int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\
&= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\
&= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C
\end{align*}
\end{questionparts}

\begin{align*}
\int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\
&= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\
&= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\
&= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\
&= \frac{\pi-\alpha}{2}
\end{align*}