Animated Solution for Mathematics - Definite Integration: If $f$ is an even function then prove that ${\int }_0^{\pi /2}f(\cos 2x)\cos xdx=\sqrt{2}{\int }_0^{\pi /4}f(\sin 2x)\cos xdx$.

Welcome, fellow traveler of the calculus realm. Today, we are embarking on a journey through the elegant landscape of definite integration.

We are tasked with proving the following identity, given that $f$ is an even function:

Unlocking the Gate

Whenever you see limits from $0$ to $a$, your mind should immediately jump to the King's Property:

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

Simplifying this expression, we obtain:

The Hidden Key

Now, look closely at $\cos (\pi -2x)$. In the second quadrant, cosine is negative, so this becomes $-\cos 2x$.

Our integral is now:

Here is where the problem statement shines: $f$ is an even function. This means $f(-u)=f(u)$, so the negative sign vanishes:

We now have two expressions for $I$: the original one with $\cos x$ and our new one with $\sin x$. Adding them gives:

This integrand is symmetric about $\frac{\pi }{4}$. By the half-interval property, we can write:

This simplifies to:

We use the trigonometric identity $\sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})$. Substituting this, we get:

By setting $u=x+\frac{\pi }{4}$, we shift our limits. Furthermore, using the identity $\cos 2x=\sin (2(\frac{\pi }{4}-x))$, we arrive at the target:

We have successfully navigated the complexity and arrived at the proof. Remember, in JEE Advanced, it is not just about the calculation; it is about seeing the symmetry.