If, $\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{\mathrm{e}}|\mathrm{f}(\theta)|+\mathrm{c}$ (where c is a constant of integration), then the ordered pair $(\lambda,|f(\theta)|)$ is equal to
$(1,|1+\tan \theta \mid)$
$(1,|1-\tan \theta|)$
$\quad(-1,|1+\tan \theta|)$
$\quad(-1,|1-\tan \theta|)$
Solution
Let
$\begin{aligned}
I & =\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)} \\
& =\int \frac{\sec ^2 \theta d \theta}{\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)+\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)} \\
& =\int \frac{\sec ^2 \theta\left(1-\tan ^2 \theta\right) d \theta}{(1+\tan \theta)^2}
\end{aligned}$
$\begin{aligned} & \text {Put } \tan \theta=t \Rightarrow \sec ^2 \theta d \theta=d t \\ \therefore \quad I & =\int \frac{\left(1-t^2\right) d t}{(1+t)^2}=\int \frac{1-t}{1+t} d t \\ & =\int \frac{2-(1+t)}{1+t} d t \\ & =2 \log |1+t|-t+C\end{aligned}$
$=2 \log |1+\tan \theta|-\tan \theta+C$ Comparing with $\lambda \tan \theta+2 \log |f(\theta)|+C$, we get
$\therefore \quad \lambda=-1, \mathrm{f}(\theta)=1+\tan \theta$