Q.$\int \dfrac{dx}{\sin(x-a)\sin(x-b)}$ is equal to
(A) $\sin(b-a)\,\log\left|\dfrac{\sin(x-b)}{\sin(x-a)}\right| + C$
(B) $\operatorname{cosec}(b-a)\,\log\left|\dfrac{\sin(x-a)}{\sin(x-b)}\right| + C$
(C) $\operatorname{cosec}(b-a)\,\log\left|\dfrac{\sin(x-b)}{\sin(x-a)}\right| + C$
(D) $\sin(b-a)\,\log\left|\dfrac{\sin(x-a)}{\sin(x-b)}\right| + C$

Concept: U Substitution — rewrite the integrand using the identity $\sin(x-a) = \sin\big((x-b)+(b-a)\big)$ and expand.

Step 1: Write

$\sin(x-a) = \sin\big((x-b)+(b-a)\big) = \sin(x-b)\cos(b-a) + \cos(x-b)\sin(b-a)$.

Step 2: The integral becomes

$\int \frac{dx}{\sin(x-a)\sin(x-b)} = \int \frac{dx}{\sin(x-b)\big[\sin(x-b)\cos(b-a) + \cos(x-b)\sin(b-a)\big]}$.

Step 3: Divide numerator and denominator by $\sin^2(x-b)$:

$= \int \frac{\csc^2(x-b)\,dx}{\cos(b-a) + \sin(b-a)\cot(x-b)}$.

Step 4: Let $t = \cot(x-b)$, so $dt = -\csc^2(x-b)\,dx$. Then

$= -\int \frac{dt}{\cos(b-a) + \sin(b-a)\,t} = -\frac{1}{\sin(b-a)}\log\big|\cos(b-a) + \sin(b-a)\cot(x-b)\big| + C$. …

We want to integrate $\int \frac{dx}{\sin(x-a)\sin(x-b)}$. The denominator is a product of two sine functions with different phase shifts. There’s no obvious direct substitution, but we can use a clever trick: introduce a constant difference in the numerator using the sine of the difference of the two angles.

Notice that $(x-a) - (x-b) = b-a$, a constant. So $\sin(b-a) = \sin[(x-a)-(x-b)]$. Expanding this using the sine subtraction formula:

This expression has exactly the same denominator terms $\sin(x-a)$ and $\sin(x-b)$ in the product. If we divide both sides by $\sin(x-a)\sin(x-b)$, we get:

That’s the central insight: the constant $\sin(b-a)$ lets us rewrite the reciprocal product as a difference of cotangents.

Now the integral becomes:

We integrate each cotangent. Recall that $\int \cot u \, du = \log|\sin u| + C$. So:

Thus:

Combine the logs:

Since $\frac{1}{\sin(b-a)} = \operatorname{cosec}(b-a)$, the final result is: …

Method: Splitting $\frac{1}{\sin(x-a)\sin(x-b)}$ with a constant angle

For a reciprocal product of two sines whose arguments differ by a constant, manufacture that constant to break the fraction into cotangents.

Steps

Step 1: Insert the constant $\sin(b-a)$.

Note $(x-a)-(x-b)=b-a$, so $\sin(b-a)=\sin\big[(x-a)-(x-b)\big]=\sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b)$.

Step 2: Divide by the product to get cotangents. …

Common Mistakes

Mistake 1: Reversing the ratio in the final logarithm.

Why it's wrong: $\cot(x-b)-\cot(x-a)$ integrates to $\log\left|\frac{\sin(x-b)}{\sin(x-a)}\right|$; swapping numerator and denominator gives the negative — the wrong option (B). Correct approach: keep the subtraction order consistent.

Mistake 2: Confusing $\operatorname{cosec}(b-a)$ with $\sin(b-a)$. …