Animated Solution for Mathematics - Indefinite Integration: For $x^2\ne n\pi +1$, $n\in N$ (the set of natural numbers), the integral $\int \sqrt{\frac{2\sin (x^2-1)-\sin 2(x^2-1)}{2\sin (x^2-1)+\sin 2(x^2-1)}}dx$ is equal to : (where c is a constant of integration)

When you first see an integral like

your heart might skip a beat. It is natural to feel intimidated.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for a hidden, elegant simplicity. Our job is not to fight the complexity, but to peel back the layers until the core beauty of the math reveals itself.

The first thing we must do is stop looking at the square root. It is a distraction. Instead, focus entirely on the fraction inside.

We see $\sin (x^2-1)$ and $\sin 2(x^2-1)$. The angle in the second term is exactly double the angle in the first. This is a classic signal to deploy our double-angle identities.

Recall that $\sin 2\theta =2\sin \theta \cos \theta$. By setting $\theta =x^2-1$, we can rewrite the entire expression:

Suddenly, the 'nightmare' is starting to look like a very organized algebraic expression. We can factor out $2\sin (x^2-1)$ from both the top and the bottom. When we cancel these terms, we are left with the much cleaner fraction:

Now, we are left with $\sqrt{\frac{1-\cos (x^2-1)}{1+\cos (x^2-1)}}$. If you have been practicing your trigonometry, your brain should immediately scream 'half-angle identities!'

We know that $1-\cos \theta =2{\sin }^2(\frac{\theta }{2})$ and $1+\cos \theta =2{\cos }^2(\frac{\theta }{2})$. Applying this to our expression, the numerator becomes $2{\sin }^2(\frac{x^2-1}{2})$ and the denominator becomes $2{\cos }^2(\frac{x^2-1}{2})$.

The twos cancel out, and we are left with $\sqrt{{\tan }^2(\frac{x^2-1}{2})}$. The square root and the square cancel out, leaving us with $\tan (\frac{x^2-1}{2})$. We have successfully tamed the beast!

Now, we address the integral. To make this solvable, we consider the differential $xdx$ present in the context of the chain rule. We are looking at:

This is where the substitution method shines. Let $t=\frac{x^2-1}{2}$. When we differentiate $t$ with respect to $x$, we get $\frac{dt}{dx}=x$, which means $dt=xdx$.

This is perfect! Our integral transforms into:

We know the integral of $\tan (t)$ is $\ln |\sec (t)|+c$. Substituting back our value for $t$, we arrive at our final answer:

Look at that. We started with a massive, intimidating expression, and through the systematic application of identities and substitution, we arrived at a concise, elegant result. This is the essence of JEE mathematics.