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: Understanding the Question:
We are given an indefinite integral problem where the result matches a specific algebraic form. We need to find the constant coefficient values $A$ and $B$.

Step 2: Key Formula or Approach:
To evaluate the integral, we modify the argument of the numerator's sine function to match the denominator by adding and subtracting $\alpha$: $x = (x - \alpha) + \alpha$. Then, we apply the trigonometric identity: $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$

Step 3: Detailed Explanation:
Let the given integral be $I$: $$I = \int \frac{\sin x}{\sin (x-\alpha)} dx$$ Rewrite the numerator argument: $$I = \int \frac{\sin [(x - \alpha) + \alpha]}{\sin (x-\alpha)} dx$$ Expand the numerator using the identity for $\sin(A+B)$: $$I = \int \frac{\sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha}{\sin(x-\alpha)} dx$$ Split the expression into two separate fractional parts: $$I = \int \left( \frac{\sin(x-\alpha)\cos\alpha}{\sin(x-\alpha)} + \frac{\cos(x-\alpha)\sin\alpha}{\sin(x-\alpha)} \right) dx$$ $$I = \int \left( \cos\alpha + \sin\alpha \cdot \cot(x-\alpha) \right) dx$$ Since $\alpha$ is a constant with respect to $x$, both $\cos\alpha$ and $\sin\alpha$ can be pulled outside the integrals: $$I = \cos\alpha \int dx + \sin\alpha \int \cot(x-\alpha) dx$$ Integrate each term step-by-step ($\int \cot g(x) dx = \log|\sin g(x)|$): $$I = (\cos\alpha)x + (\sin\alpha)\log|\sin(x-\alpha)| + c$$ Comparing this resulting equation to the standard expression provided in the prompt: $$Ax + B \log |\sin (x-\alpha)| + c$$ By matching the matching corresponding coefficients, we find: $$A = \cos\alpha \quad \text{and} \quad B = \sin\alpha$$

Step 4: Final Answer:
The values of $A$ and $B$ are $\cos \alpha, \sin \alpha$ respectively, which corresponds to option (A).