If $f(x)$ is an integrable function on $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ and $I_1=\int\limits_{\pi / 6}^{\pi / 3} \sec ^2 \theta f(2 \sin 2 \theta) d \theta$ and $I_2=\int\limits_{\pi / 6}^{\pi / 3} cosec^2 \theta f(2 \sin 2 \theta) d \theta$, then
Answer & explanation
Correct answer: option 4
Using $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$, we obtain
$I_1 =\int\limits_{\pi / 6}^{\pi / 3} \sec ^2\left(\frac{\pi}{2}-\theta\right) f(2 \sin (\pi-2 \theta)) d \theta$
$\Rightarrow I_1 =\int\limits_{\pi / 6}^{\pi / 3} cosec^2 \theta f(2 \sin 2 \theta) d \theta=I_2$