Animated Solution for Mathematics - Indefinite Integration: Let $I(x)=\int \frac{6}{{\sin }^{2}x(1-\cot x{)}^2}dx$. If $I(0)=3$, then $I(\frac{\pi }{12})$ is equal to

Welcome, future engineers. Today, we are not just solving an integral; we are embarking on a journey of pattern recognition and mathematical elegance.

When you first look at the problem it is natural to feel a slight hesitation. The expression looks cluttered, almost chaotic. But in the world of JEE Advanced, chaos is often just order in disguise.

Every great integration begins with a moment of observation. We see a ${\sin }^{2}x$ in the denominator and a $(1-\cot x{)}^2$ term. Your mathematical intuition should immediately scream: "Trigonometric Identity!"

We know that $\frac{1}{{\sin }^{2}x}$ is the reciprocal of ${\sin }^{2}x$, which is ${\csc }^{2}x$. By rewriting the integral as the problem suddenly shifts from a confusing mess to a beautiful, structured form.

We have created a relationship between the numerator and the denominator. We have a function, $(1-\cot x)$, and its derivative, ${\csc }^{2}x$, sitting right there in the numerator. This is the hallmark of a perfect substitution.

Now, let us perform the substitution. We set $t=1-\cot x$.

When we differentiate both sides with respect to $x$, we get $\frac{dt}{dx}=0-(-{\csc }^{2}x)$, which simplifies beautifully to $\frac{dt}{dx}={\csc }^{2}x$. This means $dt={\csc }^{2}xdx$.

Our integral transforms into This is the moment where the tension breaks. We have moved from the realm of complex trigonometry into the realm of simple algebraic power rules.

Applying the power rule $\int t^{n}dt=\frac{t^{n+1}}{n+1}$, we find that the integral of $t^{-2}$ is $\frac{t^{-1}}{-1}$. Multiplying by our constant $6$, we arrive at

Do not forget the constant of integration, $C$! It is the ghost in the machine, the piece of information that defines the specific curve among the family of curves. We then substitute back to get

The problem states $I(0)=3$. If you try to plug $x=0$ directly into $\cot x$, you will find it is undefined. This is the "Trap" I warned you about.

We rewrite $\cot x$ as $\frac{\cos x}{\sin x}$ to get

Now, as $x\to 0$, $\sin x\to 0$ and $\cos x\to 1$. The expression becomes $\frac{-6(0)}{0-1}+C$, which is simply $0+C$. Since we are given $I(0)=3$, we immediately find that $C=3$.

Our specific function is $I(x)=3-\frac{6}{1-\cot x}$. Finally, we evaluate at $x=\frac{\pi }{12}$.

We know that $\cot (\frac{\pi }{12})=2+\sqrt{3}$. Substituting this in, we get

To finish, we rationalize the denominator by multiplying by the conjugate $(\sqrt{3}-1)$. The denominator becomes $(\sqrt{3}{)}^2-1^2=2$.

Thus, we have The $3$ and $-3$ cancel out, leaving us with the elegant final answer: $3\sqrt{3}$.