Animated Solution for Mathematics - Definite Integration: The integral ${\int }_0^{\pi }\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx$ equals:

Conquering the Trigonometric Integral

My dear student, welcome to another deep dive into the beauty of calculus. Today, we are going to tackle an integral that looks like a monster but is actually a masterpiece of algebraic symmetry.

We are looking at the integral:

When you first see this, it is natural to feel a bit of hesitation. The square root, the trigonometric functions, and the fraction inside the argument are designed to intimidate. But in JEE Advanced, intimidation is just a mask for a simple, elegant truth waiting to be uncovered.

The Hidden Identity

Let us look at the expression inside the square root: $1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}$. Whenever you see a quadratic-like expression in a JEE problem, your first instinct should be to check for a perfect square.

Let us rearrange the terms:

Does this look familiar? It is the classic expansion of $(a-b{)}^2=a^2-2ab+b^2$. If we set $a=1$ and $b=2\sin \frac{x}{2}$, then $a^2=1$, $b^2=4{\sin }^2\frac{x}{2}$, and $-2ab=-2(1)(2\sin \frac{x}{2})=-4\sin \frac{x}{2}$.

It matches perfectly! The entire expression inside the square root is simply $(1-2\sin \frac{x}{2}{)}^2$. Our integral now looks much friendlier:

The Modulus Trap

Here is where the battle is won or lost. Many students will instinctively cancel the square root with the square and write $1-2\sin \frac{x}{2}$.

But stop! Remember the golden rule of algebra: $\sqrt{u^2}=|u|$. The square root must always yield a non-negative result.

If we ignore the modulus, we are assuming that $1-2\sin \frac{x}{2}$ is always positive, which is not necessarily true. Thus, our integral is actually: This modulus is the guardian of our mathematical integrity.

The Critical Point

To evaluate this, we need to know when the expression inside the modulus is positive and when it is negative. We find the critical point by setting $1-2\sin \frac{x}{2}=0$, which leads to $\sin \frac{x}{2}=\frac{1}{2}$.

In the interval $[0,\pi ]$, the angle $\frac{x}{2}$ ranges from $0$ to $\frac{\pi }{2}$. In this range, $\sin \frac{x}{2}=\frac{1}{2}$ occurs only at $\frac{x}{2}=\frac{\pi }{6}$, meaning $x=\frac{\pi }{3}$.

This is our turning tide. For $x\in [0,\frac{\pi }{3}]$, $\sin \frac{x}{2}<\frac{1}{2}$, so $1-2\sin \frac{x}{2}>0$. For $x\in [\frac{\pi }{3},\pi ]$, $\sin \frac{x}{2}>\frac{1}{2}$, so $1-2\sin \frac{x}{2}<0$.

The Final Integration

Now we split the integral:

For the first part:

For the second part:

Adding them together:

And there you have it! By respecting the modulus and carefully splitting the integral, we have arrived at the solution. Never fear the complexity; just break it down, step by step. The final answer is $4\sqrt{3}-4-\frac{\pi }{3}$.