Problem source

Let
	\[
	I=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta\, ,
	\]
	where $0<\alpha<\frac{1}{4}\pi$. Show that
	\[
	I=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1+\sin\theta\sin2\alpha}\,\mathrm{d}\theta\, ,
	\]
	and hence that
	\[
	I=\frac{\pi}{\sin^{2}2\alpha}-\cot^{2}2\alpha\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\sec^{2}\theta}{1+\cos^{2}2\alpha\tan^{2}\theta}\,\mathrm{d}\theta.
	\]
	Show that $I=\frac{1}{2}\pi\sec^{2}\alpha$, and state the value of $I$ if $\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi$.

Solution source

\begin{align*}
\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta &= \int_{u = \frac12 \pi}^{u = -\frac12 \pi} \frac{\cos^2 (-u)}{1-\sin(-u) \sin 2 \alpha} -\d u \tag{$u = -\theta$} \\ 
&= \int_{\frac12 \pi}^{-\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} -\d u \\
&= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 u}{1+\sin u \sin 2 \alpha} \d u \\
&= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta \\
\end{align*}

Since $\displaystyle \frac{1}{(1-a^2u^2)} = \frac12 \l \frac{1}{1+au} + \frac1{1-au} \r$

\begin{align*}
\int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1+\sin \theta \sin 2 \alpha} \d \theta  &=  \int_{-\frac12 \pi}^{\frac12 \pi} \frac{\cos^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\
&= \int_{-\frac12 \pi}^{\frac12 \pi} \frac{1-\sin^2 \theta}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\
&=  \int_{-\frac12 \pi}^{\frac12 \pi} \frac{(1-\sin ^2\theta \sin^2 2 \alpha) \frac{1}{\sin^2 2\alpha} + 1 - \cosec^2 2\alpha}{1-\sin ^2\theta \sin^2 2 \alpha} \d \theta \\
&= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta \sin^2 2 \alpha} \d \theta \\
&= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{1 - \sin^2 \theta (1-\cos^2 2 \alpha)} \d \theta \\
&= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{1}{\cos^2 \theta +\sin^2 \theta \cos^2 2 \alpha} \d \theta \\
&= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha \int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta \\
\end{align*}

Finally, using the substitution $u =|\cos 2 \alpha | \tan \theta, \d u = |\cos 2 \alpha |\sec^2 \theta \d \theta$

\begin{align*}
\int_{-\frac{\pi}2}^{\frac{\pi}2} \frac{\sec^2 \theta}{1 +\tan^2 \theta \cos^2 2 \alpha} \d \theta &= |\sec 2\alpha|\int_{u = -\infty}^{u = \infty} \frac{1}{1 + u^2} \d u  \\
&= |\sec 2 \alpha|\pi
\end{align*}

and so 

\begin{align*}
I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha  |\sec 2 \alpha|\pi \\
&= \frac{\pi}{\sin^2 2\alpha} \l 1-\cos 2\alpha \r \\
&= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \sin^2 \alpha \r \\
&= \frac{\pi}{2 \cos^2 \alpha} = \frac{\pi}{2} \sec^2 \alpha
\end{align*}

When $\alpha$ small enough that the modulus doesn't flip the sign. When if $\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi$ we have:

\begin{align*}
I &= \frac{\pi}{\sin^2 2\alpha} -\cot^2 2\alpha  |\sec 2 \alpha|\pi \\
&= \frac{\pi}{\sin^2 2\alpha} \l 1+\cos 2\alpha \r \\
&= \frac{\pi}{4\sin^2 \alpha\cos^2 \alpha} \l 2 \cos^2 \alpha \r \\
&= \frac{\pi}{2 \sin^2 \alpha} = \frac{\pi}{2} \cosec^2 \alpha
\end{align*}