Animated Solution for Mathematics - Indefinite Integration: The value of $\sqrt{2}\int \frac{\sin xdx}{\sin \left(x-\frac{\pi }{4}\right)}$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect an integral that, at first glance, might seem like a tangled mess of trigonometric functions.

We are looking at the expression $I=\sqrt{2}\int \frac{\sin x}{\sin \left(x-\frac{\pi }{4}\right)}dx$.

When you see a problem like this, it is natural to feel a bit intimidated. The numerator is simple, but the denominator is 'shifted' by $\frac{\pi }{4}$, creating a mismatch that prevents us from using standard identities immediately. But fear not—this is exactly where the beauty of calculus begins.

The Power of Substitution

In physics and mathematics, when a coordinate system or an argument is shifted, the most elegant solution is to shift your perspective along with it. We want the denominator to be as simple as possible.

Let us define a new variable, $t=x-\frac{\pi }{4}$. This immediately tells us that $x=t+\frac{\pi }{4}$, and since the derivative of a constant is zero, $dx=dt$.

By making this substitution, we transform our integral into:

Suddenly, the denominator is just $\sin t$. We have successfully cleared the fog.

Now, we face the numerator, which is a sine of a sum. This is a classic invitation to use the addition identity: $\sin (A+B)=\sin A\cos B+\cos A\sin B$.

Setting $A=t$ and $B=\frac{\pi }{4}$, we expand the numerator:

The Elegant Cancellation

Now, let us look at the values we are working with. We know that $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$.

Substituting these into our integral, we get:

Do you see it? The $\frac{1}{\sqrt{2}}$ is common to both terms in the numerator. We can factor it out, and it will meet the $\sqrt{2}$ waiting outside the integral.

They multiply to become $1$. This is the moment of clarity—the problem collapses into something remarkably simple:

The Final Integration

We are now in the home stretch. We have a sum in the numerator divided by a single term in the denominator.

We can split this into two separate integrals:

This is a standard integral. The integral of $1$ with respect to $t$ is simply $t$, and the integral of $\cot t$ is $\ln |\sin t|$.

Thus, we arrive at:

Finally, we must return to our original variable $x$. Substituting $t=x-\frac{\pi }{4}$ back into the expression, we get $I=(x-\frac{\pi }{4})+\ln |\sin (x-\frac{\pi }{4})|+C$.

Since $-\frac{\pi }{4}$ is just a constant, we merge it into our constant of integration $C$ to define a new constant $C^{\prime }$.

The final result is:

This journey shows us that even the most daunting problems are just a series of small, logical steps. By simplifying the argument, applying identities, and trusting the algebra, we turn complexity into clarity. Keep practicing, keep questioning, and most importantly, keep finding the beauty in the math.