Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{\pi }xf(\sin x)dx=A{\int }_0^{\pi /2}f(\sin x)dx$, then $A$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, looks like a standard calculus exercise, but is actually a beautiful demonstration of symmetry and elegance.

We are tasked with evaluating the integral $I={\int }_0^{\pi }xf(\sin x)dx$ and finding the constant $A$ such that $I=A{\int }_0^{\pi /2}f(\sin x)dx$.

When you look at the integral $I={\int }_0^{\pi }xf(\sin x)dx$, you see a product of two functions: an algebraic $x$ and a composite trigonometric function $f(\sin x)$.

That $x$ sitting in the front is the 'troublemaker.' It is the barrier preventing us from integrating directly.

In the world of competitive exams, whenever you see an unwanted $x$ multiplying a function in a definite integral, your intuition should immediately pivot to the most powerful tool in your arsenal: the King's Property.

The King's Property is a fundamental theorem in definite integration:

Conceptually, this property tells us that the total area under a curve remains invariant if we reflect the function across the midpoint of the interval. It is like looking at the same landscape from the opposite side—the view changes, but the terrain remains the same.

Applying this to our integral, we replace $x$ with $(\pi -x)$. Our integral becomes:

Now, we use the trigonometric identity $\sin (\pi -x)=\sin x$. Because $\sin x$ is positive in the second quadrant, the function $f(\sin x)$ remains unchanged.

This is the magic! Our integral is now $I={\int }_0^{\pi }(\pi -x)f(\sin x)dx$.

If we distribute $f(\sin x)$ into the parentheses, we get:

Look closely at the second term on the right. It is exactly our original integral $I$! We have successfully created a recursive equation:

By adding $I$ to both sides, we get $2I=\pi {\int }_0^{\pi }f(\sin x)dx$, or:

The 'troublemaker' $x$ has vanished, leaving us with a much cleaner expression.

We are almost there, but the question asks for an integral from $0$ to $\pi /2$, while we have an integral from $0$ to $\pi$. This is where the Queen's Property comes in.

It states that ${\int }_0^{2a}g(x)dx=2{\int }_0^{a}g(x)dx$ if $g(2a-x)=g(x)$.

We already verified that $f(\sin (\pi -x))=f(\sin x)$. This symmetry means the area from $0$ to $\pi /2$ is identical to the area from $\pi /2$ to $\pi$.

Thus, we have:

Substituting this back into our equation for $I$:

Comparing this to the original form $A{\int }_0^{\pi /2}f(\sin x)dx$, we find that $A=\pi$.

This problem is a masterclass in why we study properties of integrals. We didn't need to know the specific form of $f$; we only needed to understand the symmetry of the interval and the function.

Keep practicing these transformations, and soon, you won't just be solving problems—you will be seeing the hidden symmetries of the mathematical universe.