Question

Let $f$ be a integrable function over $[0,a]$ for any real value of $a$. If $I_1=\underset{0}{\overset{\pi /2}{\int }}\cos \theta f(\sin \theta +{\cos }^2\theta )d\theta$ and $I_2=\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta f(\sin \theta +{\cos }^2\theta )d\theta$, then the value of $\frac{I_1+I_2}{I_1}$ is

Answer 1

Given: $I_1=\underset{0}{\overset{\pi /2}{\int }}\cos \theta f(\sin \theta +{\cos }^2\theta )d\theta$ and $I_2=\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta f(\sin \theta +{\cos }^2\theta )d\theta$
Now,
$I_1-I_2=\underset{0}{\overset{\pi /2}{\int }}[\cos \theta -\sin 2\theta ]f(\sin \theta +{\cos }^2\theta )d\theta$
Now, taking $\sin \theta +{\cos }^2\theta =t$
$\Rightarrow I_1-I_2=\underset{1}{\overset{1}{\int }}f\left(t\right)dt=0\Rightarrow I_1=I_2\therefore \frac{I_1+I_2}{I_1}=2$