Problem source

Let 
\[
I = \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta  
\text{  and  }
J = \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta
\]
where $0 < \alpha < \frac14\pi\,$.
\begin{questionparts}
\item Show that \[
I = \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \d\theta \] and hence that \[ \displaystyle 2I = \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {2}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \]
\item Find $J$.
\item By considering $I\sin^2 2\alpha +J\cos^2 2\alpha $, or otherwise, show that  $I =\frac12 \pi \sec^2\alpha$.
\item Evaluate $I$ in the case  $\frac14\pi < \alpha < \frac12\pi$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& I &= \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta  \\
\phi = -\theta, \d \phi = - \d \theta: &&&= \int_{\phi=\frac12\pi}^{\phi=-\frac12\pi} \frac{\cos^2(-\phi)}{1-\sin(-\phi)\sin 2 \alpha} (-1) \d \phi \\
&&&=  \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\phi}{1+\sin\phi\sin2\alpha} \d\phi \\
\\
&& 2I &= \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} \, \d\theta +\int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {\cos^2\theta}{1-\sin\theta\sin2\alpha} +\frac {\cos^2\theta}{1+\sin\theta\sin2\alpha} \right) \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {2\cos^2\theta}{1-\sin^2\theta\sin^22\alpha} \right) \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {2\cos^2\theta}{1-(1-\cos^2\theta)(1-\cos^22\alpha)} \right) \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {2\cos^2\theta}{\cos^2\theta+\cos^22\alpha-\cos^2 \theta \cos^2 2 \alpha)} \right) \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {2}{1+\cos^22\alpha(\sec^2 \theta - 1))} \right) \, \d\theta \\
&&&= \int_{-\frac12 \pi}^{\frac12\pi} \left ( 
\frac {2}{1+\tan^2 \theta \cos^22\alpha} \right) \, \d\theta \\
\end{align*}

\item $\,$ \begin{align*}
&& J &=  \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\
&&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ 
&&&= \sec(2\alpha)\pi = \frac{\pi}{\cos 2 \alpha}
\end{align*}

\item $\,$ \begin{align*}
&& I\sin^2 2\alpha +J\cos^2 2\alpha &= \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha \sec^2 \theta}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\
&&&=  \int_{-\frac12\pi}^{\frac12 \pi} \frac{\sin^2 2 \alpha+\cos^2 2 \alpha (1 + \tan^2 \theta)}{1+\tan^2 \theta \cos^2 2\alpha} \d \theta \\
&&&= \pi \\
\\
\Rightarrow && I &= \frac{\pi - \pi \cos 2 \alpha}{\sin^2 2 \alpha} \\
&&&= \pi \frac{2\sin^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha} \\
&&&= \frac12 \pi \sec^2 \alpha
\end{align*}

\item If $\frac14 \pi < \alpha < \frac12 \pi$ then our calculation for $J$ is not correct.

\begin{align*}
&& J &=  \int_{-\frac12 \pi}^{\frac12\pi} 
\frac {\sec^2\theta}{1+\tan^2\theta\cos^22\alpha} \, \d\theta \\
&&&= \left [\sec 2 \alpha \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) \right]_{-\frac12 \pi}^{\frac12\pi} \\ 
&&&= \sec(2\alpha) \left ( \lim_{\theta \to \frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right) - \lim_{\theta \to -\frac{\pi}{2}} \tan^{-1} \left ( \cos 2 \alpha \tan \theta \right)  \right) \\
&&&= \sec(2\alpha) \left ( \tan^{-1} \left ( \lim_{x\to -\infty}  x \right) - \tan^{-1} \left ( \lim_{x\to \infty}  x \right)  \right) \\
&&&= -\pi \sec 2 \alpha
\end{align*}

Still using the same logic, we can say

\begin{align*}
&& I &= \frac{\pi+\pi\cos 2 \alpha}{\sin^2 2 \alpha} \\
&&&= \pi \frac{2 \cos^2 \alpha}{4 \sin^2 \alpha \cos^2 \alpha}\\
&&&= \frac12 \pi \cosec^2 \alpha
\end{align*}



\end{questionparts}