Animated Solution for Mathematics - Indefinite Integration: If $\int \sqrt{\sec 2x-1}dx=\alpha {\log }_e\left|\cos 2x+\beta +\sqrt{\cos 2x(1+\cos \frac{1}{\beta }x)}\right|+\text{constant}$, then $\beta -\alpha$ is equal to ______.

Conquering the Intimidating Integral

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions and nested radicals.

You see an integral like $I=\int \sqrt{\sec 2x-1}dx$, and your instinct might be to panic. But I want you to take a deep breath. In calculus, especially at this level, intimidation is the first trap. The problem isn't trying to defeat you; it is inviting you to simplify it.

The Trigonometric Dance

Our first step is to strip away the complexity. We see $\sec 2x$, and we immediately know its identity: $\sec 2x=\frac{1}{\cos 2x}$. Let us rewrite our integral:

By taking the lowest common multiple inside the square root, we get:

Now, look at that numerator: $1-\cos 2x$. This is a classic identity that every JEE aspirant should have etched into their memory: $1-\cos 2x=2{\sin }^{2}x$. Substituting this in, our integral becomes:

Notice the elegance here. We have isolated $\sin x$ in the numerator. This is the 'spark' we were looking for. Whenever you see a $\sin x$ term, you should immediately think of a substitution involving $\cos x$.

The Substitution Strategy

We have $\sin x$ in the numerator and $\cos 2x$ in the denominator. To make this work, we need the denominator to be in terms of $\cos x$. We use the identity $\cos 2x=2{\cos }^{2}x-1$. Our integral transforms into:

Now, let us perform the substitution. Let $t=\cos x$. Then $dt=-\sin xdx$, which means $\sin xdx=-dt$. Substituting this, we get:

To make this match the standard form $\int \frac{dt}{\sqrt{t^2-a^2}}$, we factor out the $2$ from the square root in the denominator:

See how the $\sqrt{2}$ terms canceled out? That is the beauty of mathematics—when you follow the logical path, the complexity often resolves itself.

The Algebraic Alchemy

We are now at the standard integral form. Using the formula $\int \frac{dx}{\sqrt{x^2-a^2}}=\ln |x+\sqrt{x^2-a^2}|+C$, we get:

Substituting $t=\cos x$ back in, we have:

Now, we must match the target form provided in the question. This requires careful algebraic manipulation. We simplify the expression inside the logarithm to reach a form that mirrors the target. Through squaring the argument and using the property $\ln A=\frac{1}{2}\ln A^2$, we eventually arrive at:

Comparing this to the target form, we identify $\alpha =-\frac{1}{2}$ and $\beta =\frac{1}{2}$.

The Final Victory

The final step is simply calculating $\beta -\alpha =\frac{1}{2}-(-\frac{1}{2})=1$.

This problem was not just about integration; it was about persistence. It was about taking a terrifying expression, applying the right identities, performing a clean substitution, and then having the patience to manipulate the final result to match the required form. You have mastered the process. Keep this rigor, keep this focus, and you will conquer any problem the JEE throws at you. The final answer is 1.