Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$, where $c$ is a constant of integration, then the ordered pair $(a,b)$ is equal to :

Imagine you are standing before a complex integral, one that seems designed to intimidate. The expression is:

At first glance, the square root and the trigonometric functions might seem like a chaotic mess. But in the world of JEE Advanced, chaos is just order waiting to be discovered.

The first step in any great investigation is observation. Look at the numerator: $\cos x-\sin x$. Now, look at the denominator's core: $\sin 2x$.

If you recall your basic calculus, you might notice that $\cos x-\sin x$ is the derivative of $\sin x+\cos x$. This is our "golden key." When you see a numerator that looks like the derivative of a part of the denominator, you are on the right track.

Since we have identified the derivative relationship, we proceed with a strategic substitution. We define a new variable, $t$, such that:

When we differentiate this with respect to $x$, we get $\frac{dt}{dx}=\cos x-\sin x$. Rearranging this gives us $dt=(\cos x-\sin x)dx$.

Just like that, the entire numerator of our integral is replaced by a simple $dt$. The complexity is beginning to melt away.

Now, we must address the denominator. We have $\sin 2x$ trapped inside a square root. We need to express this in terms of our new variable $t$.

This is where the "squaring trick" comes into play. We take our substitution $t=\sin x+\cos x$ and square both sides:

Expanding the right side, we get $t^2={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x$. Using the fundamental identity ${\sin }^{2}x+{\cos }^{2}x=1$ and the double-angle formula $2\sin x\cos x=\sin 2x$, the equation simplifies to:

We have successfully built a bridge between the original variable $x$ and our new variable $t$.

With our pieces in place, we substitute everything back into the integral:

Distributing the negative sign, we get $\int \frac{dt}{\sqrt{8-t^2+1}}$, which simplifies beautifully to:

This is a standard integral form. We can write $9$ as $3^2$, giving us $\int \frac{dt}{\sqrt{3^2-t^2}}$. The standard formula for this is:

The Elegance of the Result

Finally, we perform back-substitution, replacing $t$ with $\sin x+\cos x$. Our result is:

Comparing this to the target form $a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+C$, we immediately see that $a=1$ and $b=3$. The ordered pair $(a,b)$ is $(1,3)$.

This problem is a perfect example of how recognizing patterns can turn a daunting task into a series of logical, elegant steps. Keep practicing these derivative pairs, and you will find that even the most complex integrals become second nature.