Animated Solution for Mathematics - Definite Integration: If $f:\mathbb{R}\to \mathbb{R}$ be a continuous function satisfying ${\int }_0^{\pi /2}f(\sin 2x)\sin xdx+\alpha {\int }_0^{\pi /4}f(\cos 2x)\cos xdx=0$, then the value of $\alpha$ is

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey through the elegance of calculus.

When you first look at the equation

it is natural to feel a bit overwhelmed. You see an arbitrary function $f$, two different limits, and a mysterious constant $\alpha$. But take a deep breath; in the world of JEE Advanced, complexity is often just a mask for symmetry.

The Divide and Conquer

Our first objective is to make the two integrals speak the same language. The first integral, let us call it $I$, spans from $0$ to $\pi /2$.

The second integral spans from $0$ to $\pi /4$. The discrepancy in limits is our first clue. We must split $I$ at the point $x=\pi /4$.

By the fundamental properties of definite integrals, we can write $I$ as:

Now, the first part is already aligned with our target. The second part, however, is the one that needs our attention.

The Mirroring Principle

Look at the second integral: ${\int }_{\pi /4}^{\pi /2}f(\sin 2x)\sin xdx$. We want the limits to be $0$ to $\pi /4$.

We use the substitution $x=\pi /2-t$. This is the 'Reflection Principle.' When $x=\pi /4$, $t=\pi /4$. When $x=\pi /2$, $t=0$. The differential $dx$ becomes $-dt$.

As we substitute, $\sin 2x$ becomes $\sin (2(\pi /2-t))=\sin (\pi -2t)=\sin 2t$. Similarly, $\sin x$ becomes $\sin (\pi /2-t)=\cos t$.

After accounting for the negative sign from the differential, the limits flip back to $0$ to $\pi /4$. We have successfully transformed the second part into:

Now, both parts of $I$ share the same limits.

The Elegant Cancellation

We combine the parts:

We are close, but our target involves $f(\cos 2x)$, not $f(\sin 2x)$. We need one final, powerful substitution: $x=\pi /4-t$. This is our 'Symmetry Bridge.'

Watch what happens to the argument: $\sin 2x$ becomes $\sin (2(\pi /4-t))=\sin (\pi /2-2t)=\cos 2t$. This is exactly what we need.

Now, what about the term $(\sin x+\cos x)$? Substituting $x=\pi /4-t$ gives us $\sin (\pi /4-t)+\cos (\pi /4-t)$.

Using the compound angle formulas, this expands to:

The $\sin t$ terms vanish into thin air, leaving us with $\sqrt{2}\cos t$.

Putting it all together, our integral $I$ transforms into:

Substituting this back into our original equation, we get:

Factoring out the integral, we see that $(\sqrt{2}+\alpha )\times \text{Integral}=0$. For this to hold for any continuous function $f$, the coefficient must be zero.

Thus, $\alpha =-\sqrt{2}$. You did it! You navigated the substitution, respected the symmetry, and emerged victorious.