Animated Solution for Mathematics - Definite Integration: If $f(x)=\frac{2-x\cos x}{2+x\cos x}$ and $g(x)={\log }_{e}x,(x>0)$ then the value of integral ${\int }_{-\pi /4}^{\pi /4}g(f(x))dx$ is :

Welcome, fellow traveler on the JEE Advanced journey. Today, we are going to tackle a problem that, at first glance, looks like a nightmare of calculus.

You see an integral, you see a complex composite function, and your instinct might be to reach for your pen and start grinding through integration by parts or some terrifying substitution. But stop. Take a breath.

In the world of JEE Advanced, the most complex-looking problems often have the most elegant, simple solutions hidden right in plain sight.

Let us look at our integral:

If you try to integrate this directly, you will be lost in a forest of derivatives and anti-derivatives. But look at the limits of integration: $[-\frac{\pi }{4},\frac{\pi }{4}]$.

Whenever you see symmetric limits like this—from $-a$ to $a$—your brain should immediately sound an alarm. This is a classic hallmark of a symmetry problem. We are not meant to integrate; we are meant to observe.

Even or Odd?

To unlock the secret of this integral, we must test the function $h(x)={\log }_e\left(\frac{2-x\cos x}{2+x\cos x}\right)$ for symmetry. Is it even? Is it odd? Or is it neither?

We test this by replacing $x$ with $-x$. Let us perform this substitution carefully:

Now, we apply our trigonometric knowledge. We know that $\cos (-x)=\cos (x)$ because cosine is an even function. This is a crucial moment.

The negative sign inside the cosine vanishes, but the negative sign multiplying the $x$ remains. Our expression becomes:

Look at what we have now. The numerator and denominator have swapped places compared to our original function $h(x)$. In the world of fractions, this means we are looking at the reciprocal.

We can rewrite this as:

Now, we invoke the power rule of logarithms, which tells us that $\log (a^n)=n\log (a)$. That exponent of $-1$ jumps out to the front, and suddenly, the magic happens:

We have just proven that $h(-x)=-h(x)$. By definition, this means our function $h(x)$ is an odd function.

Why does this matter? Because for any odd function, the area under the curve from $-a$ to $0$ is the exact negative of the area from $0$ to $a$.

When you integrate an odd function over symmetric limits, the two areas cancel each other out perfectly. The integral is zero.

Finally, we look at our options. We calculated $0$, but the options are logarithmic expressions. We know that ${\log }_{e}1=0$.

Thus, option 4 is our destination. You see? We didn't need to perform a single complex integration. We just needed to observe, test, and trust the properties of functions. Keep this mindset, and you will conquer any problem the JEE throws at you.