Q.Integrate the function $\frac{\sin x}{\sin(x-a)}$

The key idea is to rewrite the numerator using the sine angle subtraction formula, so the integrand simplifies into a constant minus a cotangent term.

Step 1: Write $\sin x = \sin((x-a) + a) = \sin(x-a)\cos a + \cos(x-a)\sin a$.

Step 2: Substitute into the integrand:

$\frac{\sin x}{\sin(x-a)} = \frac{\sin(x-a)\cos a + \cos(x-a)\sin a}{\sin(x-a)} = \cos a + \sin a \cdot \frac{\cos(x-a)}{\sin(x-a)}.$ …

Why This Approach Works

When you see an integrand like $\frac{\sin x}{\sin(x-a)}$, your first instinct might be to try a direct substitution. But the denominator $\sin(x-a)$ is a shifted version of the sine function, and the numerator is just $\sin x$. The trick is to notice that $x = (x-a) + a$ — a simple shift. This lets us express $\sin x$ in terms of $\sin(x-a)$ and $\cos(x-a)$, which will cancel beautifully with the denominator.

The sine addition formula is your friend here: $\sin(A+B) = \sin A \cos B + \cos A \sin B$. By setting $A = x-a$ and $B = a$, we get $\sin x = \sin((x-a)+a) = \sin(x-a)\cos a + \cos(x-a)\sin a$. This turns a messy fraction into a sum of two simple terms.

Step-by-Step Solution

1. Rewrite the numerator using the angle addition formula. Let $u = x-a$, so $x = u + a$. Then:

The integrand becomes:

2. Simplify the resulting expression. Since $\frac{\cos u}{\sin u} = \cot u$, we have:

Now substitute back $u = x-a$:

3. Integrate term by term. The integral becomes:

The first integral is straightforward: $\int \cos a \, dx = x \cos a + C_1$.

4. Handle the cotangent integral. Recall that $\int \cot \theta \, d\theta = \log|\sin \theta| + C$. For $\cot(x-a)$, let $t = x-a$, so $dt = dx$, and:

Therefore: …

Method: Rewrite the numerator's angle as (denominator's angle) + constant

Use this for $\dfrac{\sin(\text{angle}_1)}{\sin(\text{angle}_2)}$ (or the cosine version) where the two angles differ by a constant — expanding with the addition formula splits the fraction into a constant plus a cotangent/tangent.

Steps

Step 1: Express the top angle through the bottom angle.

Write $x=(x-a)+a$, so $\sin x=\sin\bigl((x-a)+a\bigr)$ and expand with $\sin(A+B)=\sin A\cos B+\cos A\sin B$.

Step 2: Divide through by the denominator.

The $\sin(x-a)$ cancels in one term, leaving …

Common Mistakes

Mistake 1: Attempting a direct substitution $u=\sin(x-a)$.

Why it's wrong: the numerator $\sin x$ is not a multiple of $\cos(x-a)\,dx$, so the substitution fails. Correct approach: rewrite $\sin x=\sin\bigl((x-a)+a\bigr)$ and expand.

Mistake 2: Attaching $\sin a$ only to the argument, not the whole log. …