Problem source

Show that
\[
\sin\theta = \frac {2t}{1+t^2}, \ \ \
\cos\theta = \frac{1-t^2}{1+t^2}, \ \ \
\frac{1+\cos\theta}{\sin\theta}  = \tan (\tfrac{1}{2}\pi-\tfrac{1}{2}\theta),
\]
where $t =\tan\frac{1}{2}\theta$.
Use the substitution $t =\tan\frac{1}{2}\theta$ to show that, for $0<\alpha<\frac{1}{2}\pi$,
\[
\int_0^{\frac{1}{2}\pi}
{1 \over {1 + \cos\alpha \sin \theta}} \,\d\theta  
=\frac{\alpha}{\sin\alpha}\,,
\]
and deduce a similar result for
\[
\int_0^{\frac{1}{2}\pi}
{1 \over {1 + \sin\alpha \cos \theta}} \,\d\theta  \,.
\]

Solution source

\begin{align*}
&& \frac{2t}{1+t^2} &= \frac{2 \sin \tfrac12 \theta \cos\tfrac12 \theta }{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\
&&&= \frac{\sin \theta}{1} = \sin \theta \\
\\
&& \frac{1-t^2}{1+t^2} &= \frac{\cos^2 \tfrac12 \theta - \sin^2 \tfrac12 \theta}{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\
&&&= \frac{\cos \theta }{1}  = \cos \theta \\
\\
&& \tan(\tfrac12 \pi - \tfrac12 \theta) &= \frac{1}{t} \\
&& \frac{1+\cos \theta}{\sin \theta} &= \frac{1 + \frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} \\
&&&= \frac{2}{2t} = \frac1t = \tan(\tfrac12\pi - \tfrac12 \theta)
\end{align*}

Notice also that $\frac{\d t}{\d \theta} = \tfrac12 \sec^2 \tfrac12 \theta = \tfrac12(1 + t^2)$ so

\begin{align*}
&& I &= \int_0^{\frac12 \pi} \frac{1}{1 + \cos \alpha \sin \theta} \d \theta \\
t = \tan \tfrac12 \theta, \d \theta = \frac{2}{1+t^2} \d t: &&&= \int_{0}^{1} \frac{1}{1 + \cos \alpha \frac{2t}{1+t^2}}\frac{2}{1+t^2} \d t \\
&&&= \int_0^1 \frac{2}{1+t^2 + 2\cos \alpha t} \d t \\
&&&= \int_0^1 \frac{2}{(t + \cos \alpha)^2+\sin^2 \alpha} \d t \\
&&&= \left [ \frac{2}{\sin \alpha} \tan^{-1} \left ( \frac{t+ \cos \alpha}{\sin \alpha} \right) \right]_0^1 \\
&&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left ( \frac{1+ \cos \alpha}{\sin \alpha} \right) -  \tan^{-1} \left ( \frac{ \cos \alpha}{\sin \alpha} \right) \right) \\
&&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left (\tan (\tfrac12 \pi - \tfrac12 \alpha \right) -  \tan^{-1} \left (\tan(\tfrac12\pi - \alpha )\right) \right) \\
&&&= \frac{2}{\sin \alpha} \left ( \tfrac12 \pi - \tfrac12 \alpha - \tfrac12 \pi + \alpha \right) \\
&&&= \frac{\alpha}{\sin \alpha}
\end{align*}

\begin{align*}
&& J &= \int_0^{\tfrac12 \pi} \frac{1}{1 + \sin \alpha \cos \theta} \d \theta \\
&&&= \int_0^{\tfrac12 \pi} \frac{1}{1 + \cos (\tfrac12 \pi - \alpha) \sin \theta} \d \theta \\
&&&= \frac{\tfrac12 \pi - \alpha}{\cos \alpha}
\end{align*}