Animated Solution for Mathematics - Definite Integration: If $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx$, then ${\int }_0^{\frac{\pi }{2}}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$ equals :

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of trigonometric powers and an intrusive variable $x$.

In the world of JEE Advanced, when you see an integral of the form ${\int }_0^{a}xf(x)dx$, your intuition should immediately scream, "King's Property!" This is not just a formula; it is a surgical tool designed to excise the variable $x$ that prevents us from integrating directly.

Let us call our target integral:

Our mission is to evaluate this area under the curve from $0$ to $\frac{\pi }{2}$.

The King's Property states that ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$. Here, our upper limit $a$ is $\frac{\pi }{2}$.

When we apply this, we replace every $x$ with $(\frac{\pi }{2}-x)$. We know that $\sin (\frac{\pi }{2}-x)=\cos x$ and $\cos (\frac{\pi }{2}-x)=\sin x$.

When we substitute these into our integral, the denominator ${\sin }^{4}x+{\cos }^{4}x$ transforms into ${\cos }^{4}x+{\sin }^{4}x$. It remains completely unchanged!

This is the hidden geometric reality of the problem: the denominator is symmetric about the midpoint of the interval. The numerator, however, changes from $x\sin x\cos x$ to $(\frac{\pi }{2}-x)\cos x\sin x$. The "troublemaker" $x$ has been shifted, but the structure is preserved.

Now, we perform the masterstroke. We add our original integral $I_0$ to the transformed integral $I_0$. Because the denominators are identical, we can combine the numerators directly:

Look closely at the numerator. We have $x\sin x\cos x$ plus $(\frac{\pi }{2}-x)\sin x\cos x$. The $x$ terms cancel out perfectly!

We are left with:

Suddenly, the variable $x$ that made the integral impossible to solve is gone. We are left with a constant $\frac{\pi }{2}$ and a standard trigonometric integral.

We divide by $2$ to isolate $I_0$:

To solve this remaining integral, we need to bring it into the realm of algebraic substitution. We divide both the numerator and the denominator by ${\cos }^{4}x$.

This is a classic maneuver for even powers of sine and cosine. The denominator becomes ${\tan }^{4}x+1$, and the numerator becomes $\tan x{\sec }^{2}x$. Our integral now looks like this:

Now, we choose our substitution. Let $t={\tan }^{2}x$. Differentiating both sides, we get $dt=2\tan x{\sec }^{2}xdx$, which implies $\tan x{\sec }^{2}xdx=\frac{dt}{2}$.

We must also update our limits: when $x=0$, $t=0$; as $x\to \frac{\pi }{2}$, $t\to \infty$. Substituting these into our expression, we get:

We are staring at a standard integral: $\int \frac{1}{1+t^2}dt={\tan }^{-1}t$. Evaluating this from $0$ to $\infty$ is straightforward, where ${\tan }^{-1}(\infty )=\frac{\pi }{2}$ and ${\tan }^{-1}(0)=0$.

Thus, our final result is:

And there it is. Through the symmetry of the King's Property and a clever substitution, we have navigated the complexity. Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the elegant path that makes the difficulty vanish.