Question

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = Ax + B \log |\sin (x-\alpha)| + c$, then the value of A and B are respectively (where c is a constant of integration)

• A. $\cos \alpha, \sin \alpha$
• B. $\sin \alpha, \cos \alpha$
• C. $-\cos \alpha, \sin \alpha$
• D. $-\sin \alpha, \cos \alpha$

Hint:

An alternative trick to solve this without integrating is to differentiate the right-hand side with respect to $x$. Differentiating $Ax + B \log|\sin(x-\alpha)|$ gives $A + B\cot(x-\alpha)$. Setting this equal to the integrand $\frac{\sin x}{\sin(x-\alpha)}$ at $x = \alpha$ or analyzing the components quickly yields $A = \cos\alpha$ and $B = \sin\alpha$.

Answer 1

The Correct Option is A

Step 1: State the goal.
We integrate $\dfrac{\sin x}{\sin(x-\alpha)}$ and compare with $Ax + B\log|\sin(x-\alpha)| + c$ to read off $A$ and $B$.
Step 2: Rewrite the angle in the numerator.
Write $x = (x-\alpha) + \alpha$ so the numerator and denominator share the angle $(x-\alpha)$.
Step 3: Expand using the sine addition rule.
$\sin\big[(x-\alpha)+\alpha\big] = \sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha$.
Step 4: Split the fraction.
Dividing each term by $\sin(x-\alpha)$ gives the integrand $\cos\alpha + \sin\alpha \cdot \cot(x-\alpha)$.
Step 5: Integrate term by term.
Since $\alpha$ is constant, $\int \cos\alpha\,dx = (\cos\alpha)x$, and $\int \sin\alpha\,\cot(x-\alpha)\,dx = (\sin\alpha)\log|\sin(x-\alpha)|$. So the integral is $(\cos\alpha)x + (\sin\alpha)\log|\sin(x-\alpha)| + c$.
Step 6: Compare coefficients.
Matching with $Ax + B\log|\sin(x-\alpha)| + c$ gives $A = \cos\alpha$ and $B = \sin\alpha$, which is option 1 and matches the key.
\[ \boxed{A = \cos\alpha,\ B = \sin\alpha} \]