Animated Solution for Mathematics - Definite Integration: Show that ${\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx$.

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are uncovering a hidden symmetry in the fabric of calculus.

You have likely encountered the integral $I={\int }_0^{\pi }xf(\sin x)dx$ and felt a moment of hesitation. How do we handle that $x$ multiplying the function? It seems to break the simplicity of the sine function, but the $x$ is actually the key to the solution.

Imagine you are looking at the area under the curve $y=xf(\sin x)$ from $0$ to $\pi$. The King's Property is our most elegant tool. It states that for any continuous function $g(x)$, the integral ${\int }_0^{a}g(x)dx$ is identical to ${\int }_0^{a}g(a-x)dx$.

Integration is the accumulation of area. If you reverse the direction of your walk along the $x$-axis—starting from $\pi$ and moving back to $0$—the total area accumulated remains the same. Let us apply this to our integral:

By replacing $x$ with $(\pi -x)$, we obtain:

Now, look closely at the term $f(\sin (\pi -x))$. We know from the unit circle that $\sin (\pi -x)=\sin x$.

This is the moment of clarity. The function $f(\sin x)$ is perfectly symmetric about the line $x=\pi /2$. Because of this, the transformation does not distort the function at all. Our integral now looks like this:

This is where the magic happens. Let us distribute the $f(\sin x)$ inside the integral:

We can split this into two separate integrals:

Do you see it? The second term on the right is exactly our original integral $I$. We have successfully created a self-referential equation. We can now replace that integral with $I$:

We are now in the home stretch. We have an equation where $I$ appears on both sides. By adding $I$ to both sides, we get:

Dividing by $2$, we arrive at the beautiful, clean result:

Take a moment to appreciate what just happened. We did not need to know the explicit form of $f$, nor did we need to perform a grueling integration by parts.

We simply used the inherent symmetry of the interval and the function. This is the essence of JEE Advanced mathematics: finding the path of least resistance through the beauty of symmetry. Keep this King's Property in your toolkit; it will serve you well in the battles to come.