Let $f$ be an integrable function defined on $[0, a]$. If $I_1=\int\limits_0^{\pi / 2} \cos \theta f\left(\sin \theta+\cos ^2 \theta\right) d \theta$
and, $I_2=\int\limits_0^{\pi / 2} \sin 2 \theta f\left(\sin \theta+\cos ^2 \theta\right) d \theta$, then
Answer & explanation
Correct answer: option 1
We have,
$I_1-I_2=\int\limits_0^{\pi / 2}(\cos \theta-\sin 2 \theta) f\left(\sin \theta+\cos ^2 \theta\right) d \theta$
$\Rightarrow I_1-I_2=\int\limits_0^1 f\left(\sin \theta+\cos ^2 \theta\right) d\left(\sin \theta+\cos ^2 \theta\right)$
$\Rightarrow I_1-I_2=\int\limits_1^1 f(t) d t=0$, where $t=\sin \theta+\cos ^2 \theta$
$\Rightarrow I_1=I_2$