Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log \sin (x-\alpha )+C$, then value of $(A,B)$ is

We are presented with the integral:

At first glance, this looks like a standard problem, but it hides a subtle trap: the mismatch between the numerator's angle $x$ and the denominator's angle $(x-\alpha )$. This is the core of our challenge.

Why is this hard?

The difficulty arises because the variable of integration is $x$, but the denominator is locked into a specific phase shift of $\alpha$. We cannot simply integrate the numerator and denominator separately. We need a bridge to reconcile these two expressions.

The Add and Subtract Trick

We need to express the numerator in terms of the denominator. We use the identity $x=(x-\alpha )+\alpha$.

This is not just a trick; it is a transformation that aligns our perspective. Now, our integral becomes:

Using the Compound Angle Formula

We invoke the identity $\sin (A+B)=\sin A\cos B+\cos A\sin B$. With $A=(x-\alpha )$ and $B=\alpha$, we expand the numerator:

Now, we split the fraction. The first term is:

The second term is:

Bringing it Home

We integrate term by term. The integral of $\cos \alpha$ with respect to $x$ is simply $x\cos \alpha$.

The integral of $\sin \alpha \cot (x-\alpha )$ is $\sin \alpha \ln |\sin (x-\alpha )|$. Combining these, we get:

Comparing this to the given form $Ax+B\ln |\sin (x-\alpha )|+C$, we identify $A=\cos \alpha$ and $B=\sin \alpha$. You have mastered the technique of angle manipulation.