Animated Solution for Mathematics - Definite Integration: Evaluate the following: ${\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$

Welcome, fellow traveler on the road to JEE excellence. Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of trigonometric functions. We are tasked with evaluating the definite integral:

When you see an $x$ sitting in the numerator of a definite integral, your intuition should immediately scream: Symmetry! The interval $[0,\pi /2]$ is a playground for the King's Property.

Imagine you are standing before a locked door. The $x$ in the numerator is the lock. To open it, we use the King's Property: ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

By replacing $x$ with $(\frac{\pi }{2}-x)$, we don't just change the variable; we transform the very soul of the integrand. Using the identities $\sin (\frac{\pi }{2}-x)=\cos x$ and $\cos (\frac{\pi }{2}-x)=\sin x$, our integral $I$ becomes:

Now, here is the magic. If we add our original integral to this new version, the $x$ terms in the numerator cancel out beautifully: $x+(\frac{\pi }{2}-x)=\frac{\pi }{2}$. We are left with:

We now focus on solving the reduced integral ${\int }_0^{\frac{\pi }{2}}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$. To make this manageable, we divide both the numerator and the denominator by ${\cos }^{4}x$. This is a classic JEE maneuver to introduce $\tan x$ and ${\sec }^{2}x$:

Let us perform a substitution. Let $t={\tan }^{2}x$. Then, the derivative $dt=2\tan x{\sec }^{2}xdx$, which implies $\tan x{\sec }^{2}xdx=\frac{1}{2}dt$.

As $x$ goes from $0$ to $\frac{\pi }{2}$, the variable $t$ travels from $0$ to $\infty$.

Substituting these into our integral, we get:

Simplifying this expression, we find:

We know that the integral of $\frac{1}{1+t^2}$ is simply ${\tan }^{-1}(t)$. Evaluating this from $0$ to $\infty$ gives us ${\tan }^{-1}(\infty )-{\tan }^{-1}(0)=\frac{\pi }{2}-0=\frac{\pi }{2}$.

Finally, multiplying by our constant $\frac{\pi }{8}$, we arrive at the elegant result:

Take a moment to appreciate this. We started with a complex, variable-dependent integral and, through the symmetry of the King's Property and a clever substitution, distilled it down to a fundamental inverse trigonometric value.