Animated Solution for Mathematics - Definite Integration: Show that ${\int }_0^{\pi /2}f(\sin 2x)\sin xdx=\sqrt{2}{\int }_0^{\pi /4}f(\cos 2x)\cos xdx$

Welcome, future IITians. Today, we are not just solving a calculus problem; we are conducting a symphony of symmetry.

When you look at an integral like $I={\int }_0^{\frac{\pi }{2}}f(\sin 2x)\sin xdx$, it is easy to feel intimidated. The function $f$ is unknown, the limits are specific, and the structure seems rigid.

In the world of JEE Advanced, rigidity is just an invitation to find the hidden flexibility.

Our journey begins with the most powerful tool in the definite integral toolkit: the King's Property. It states that for any continuous function $g(x)$, the integral from $a$ to $b$ is invariant if we replace $x$ with $a+b-x$.

By replacing $x$ with $\frac{\pi }{2}-x$, we transform our integral into:

Let us appreciate the beauty of trigonometry here. The argument $\sin 2(\frac{\pi }{2}-x)$ simplifies to $\sin (\pi -2x)$, which is simply $\sin 2x$. Meanwhile, $\sin (\frac{\pi }{2}-x)$ becomes $\cos x$.

Suddenly, our integral has a twin:

Now, we have two expressions for $I$. In mathematics, when you have two paths to the same destination, adding them is often the key to unlocking the gate.

Let us add these two versions of $I$:

Look at this new integrand. It is symmetric about the midpoint $x=\frac{\pi }{4}$. If you replace $x$ with $\frac{\pi }{2}-x$, the function remains unchanged.

Because of this symmetry, we can invoke the property that the integral from $0$ to $2a$ of a symmetric function is twice the integral from $0$ to $a$. Here, $2a=\frac{\pi }{2}$, so $a=\frac{\pi }{4}$:

With a swift cancellation of the factor of $2$, we arrive at:

We are almost there. We need to reach the form involving $f(\cos 2x)\cos x$. The term $(\sin x+\cos x)$ is begging to be compressed.

We multiply and divide by $\sqrt{2}$ to get $\sqrt{2}\sin (x+\frac{\pi }{4})$:

Now, let us perform the final substitution. Let $t=x+\frac{\pi }{4}$. This shifts our limits from $[0,\frac{\pi }{4}]$ to $[\frac{\pi }{4},\frac{\pi }{2}]$.

The term $\sin 2x$ becomes $\sin (2t-\frac{\pi }{2})$, which is $-\cos 2t$. After one more substitution to flip the limits back to $[0,\frac{\pi }{4}]$ and a final trigonometric identity check where $f(-\cos (\pi -2y))=f(\cos 2y)$, we arrive at the destination:

We have successfully navigated the complexity. Remember, in JEE Advanced, the math is rarely about brute force; it is about recognizing the symmetry and trusting the process.