Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{\pi /4}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx=\frac{1}{a}{\log }_e(\frac{a}{3})+\frac{\pi }{b\sqrt{3}}$, where $a,b\in \mathbb{N}$, then $a+b$ is equal to

Taming the Trigonometric Beast

Welcome, fellow traveler on the JEE Advanced journey. Today, we are not just solving an integral; we are dismantling a puzzle.

When you first look at the integral it is natural to feel a moment of hesitation. It looks messy and does not belong in any standard table of integrals. But that is exactly where the beauty lies; in the world of JEE, complexity is often just a mask for a simpler, more elegant structure waiting to be uncovered.

The Linearization

The first step in any battle is to know your terrain. We have ${\sin }^{2}x$ in the numerator and a product $\sin x\cos x$ in the denominator, which are classic signs that we need to linearize.

We reach into our toolkit and pull out the double-angle identities:

By substituting these, we transform our integral into:

Multiplying the numerator and denominator by $2$ gives us a much cleaner expression:

Suddenly, the chaos has subsided. We have a rational function of $\cos 2x$ and $\sin 2x$.

Divide and Conquer

Now, look at the numerator: $1-\cos 2x$. It is begging to be split because we have two distinct mathematical personalities here.

We can write our integral as:

Let us call these $I_1$ and $I_2$. By splitting them, we have turned one impossible problem into two manageable ones. This is the essence of mathematical strategy: break the problem until the pieces are small enough to handle.

The Logarithmic Path ($I_2$)

Let us tackle $I_2={\int }_0^{\pi /4}\frac{\cos 2x}{2+\sin 2x}dx$ first. Notice the relationship between the numerator and the denominator; the derivative of $2+\sin 2x$ is $2\cos 2x$.

Our numerator is almost exactly that. By using the substitution $u=2+\sin 2x$, we get $du=2\cos 2xdx$, or $\cos 2xdx=\frac{du}{2}$.

The integral becomes:

Evaluating this from $0$ to $\pi /4$ gives us:

We have conquered the first half.

The Arctan Path ($I_1$)

Now for $I_1={\int }_0^{\pi /4}\frac{1}{2+\sin 2x}dx$. This is where many students stumble, as we have no cosine in the numerator to help us.

This is the moment to use the universal substitution $t=\tan x$. We know $\sin 2x=\frac{2t}{1+t^2}$. Substituting this, and remembering that $dx=\frac{dt}{1+t^2}$, the integral transforms into:

To solve this, we complete the square: $t^2+t+1=(t+\frac{1}{2}{)}^2+\frac{3}{4}$. Now we are in the territory of the standard integral $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}(\frac{x}{a})$.

With $a=\frac{\sqrt{3}}{2}$, the integration yields:

Plugging in the limits, we get:

We combine our results:

To match the target form $\frac{1}{a}\ln (\frac{a}{3})+\frac{\pi }{b\sqrt{3}}$, we rewrite the log term as $+\frac{1}{2}\ln (\frac{2}{3})$. Comparing the two, we find $a=2$ and $b=6$.

Thus, $a+b=8$. You see? It wasn't magic. It was just a sequence of logical, beautiful steps. Keep practicing, and soon, you will see these patterns before you even pick up your pen.