Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{{\sin }^{2}x{\cos }^{2}x}{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x{)}^2}dx$ is equal to :

Welcome, fellow JEE warriors. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare.

Consider the integral:

It is a massive, intimidating polynomial of trigonometric functions. However, in JEE Advanced, complexity is often just a mask for a hidden, elegant simplicity.

We start by isolating the denominator: $D={\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x$. If you try to integrate this as is, you will be lost in a sea of algebra.

Instead, we look for patterns. Notice the first and third terms: ${\sin }^{5}x+{\sin }^{3}x{\cos }^{2}x$. If we factor out ${\sin }^{3}x$, we get:

Similarly, for the second and fourth terms, ${\cos }^{3}x{\sin }^{2}x+{\cos }^{5}x$, we factor out ${\cos }^{3}x$ to get:

The entire denominator collapses from a four-term monster into $({\sin }^{3}x+{\cos }^{3}x{)}^2$. This is our first major victory.

Now our integral simplifies to:

To transform this into a standard form, we convert the expression into $\tan x$ and $\sec x$ by dividing the numerator and denominator by ${\cos }^{6}x$.

In the numerator, we get:

In the denominator, we bring the ${\cos }^{6}x$ inside the square as ${\cos }^{3}x$:

The integral is now reduced to:

Look at the transformed integral; it is perfectly set up for substitution. Let $t=1+{\tan }^{3}x$.

Then, the differential is $dt=3{\tan }^{2}x{\sec }^{2}xdx$, which implies ${\tan }^{2}x{\sec }^{2}xdx=\frac{1}{3}dt$. Our integral becomes:

This is a basic power rule integration. The result is:

Substituting back $t=1+{\tan }^{3}x$, we obtain the final result:

We have successfully dismantled the monster. Remember, in JEE, never rush to calculate; always pause, look for patterns, and simplify.