JAMIA 2022 — Mathematics PYQ

Step 1: Identify the Function

Let $f(x)=\log \left(\frac{2-\sin x}{2+\sin x}\right)$.

To check if the function is even or odd, we replace $x$ with $-x$:

$f(-x)=\log \left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)$

Since $\sin (-x)=-\sin x$, the expression becomes:

$f(-x)=\log \left(\frac{2+\sin x}{2-\sin x}\right)$

Step 2: Apply Logarithmic Properties

Using the property $\log (\frac{a}{b})=-\log (\frac{b}{a})$, we can rewrite $f(-x)$:

$f(-x)=\log \left({\left(\frac{2-\sin x}{2+\sin x}\right)}^{-1}\right)$

$f(-x)=-\log \left(\frac{2-\sin x}{2+\sin x}\right)$

$f(-x)=-f(x)$

Step 3: Determine the Integral Value

Since $f(-x)=-f(x)$, the function $f(x)$ is an odd function.

According to the property of definite integrals for an odd function over a symmetric interval $[-a,a]$:

${\int }_{-a}^{a}f(x)dx=0$

In this case, $a=\pi /2$. Therefore:

$I={\int }_{-\pi /2}^{\pi /2}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx=0$

Final Answer:

$I=0$