Question

If $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$, then $I$ is given by}

• A. $\frac{1}{\sin(b-a)} \log |\sin(x-a) \sin(x-b)| + c$, where $c$ is a constant of integration.
• B. $\log \left|\frac{\sin(x-a)}{\sin(x-b)}\right| + c$, where $c$ is a constant of integration.
• C. $\frac{1}{\sin(b-a)} \log \left|\frac{\sin(x-a)}{\sin(x-b)}\right| + c$, where $c$ is a constant of integration.
• D. $\frac{1}{\sin(b-a)} \log \left|\frac{\sin(x-b)}{\sin(x-a)}\right| + c$, where $c$ is a constant of integration.

Hint:

For integrals of the form $\int \frac{dx}{\sin(x-a)\sin(x-b)}$ or similar products of sines/cosines in the denominator, multiply the numerator and denominator by $\sin(b-a)$ (or $\cos(b-a)$ if appropriate) and then use trigonometric sum/difference formulas to split the fraction.

Answer 1

The Correct Option is D

Step 1: Identify the form of the integral. We need to evaluate $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$.
Step 2: Introduce a constant factor to the numerator. Multiply and divide by $\sin(b-a)$. This is a common technique for this type of integral. \[ I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\sin(x-a) \sin(x-b)} dx \]
Step 3: Express $(b-a)$ in terms of $(x-a)$ and $(x-b)$. Notice that $(b-a) = (x-a) - (x-b)$. Substitute this into the numerator. \[ I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{\sin(x-a) \sin(x-b)} dx \]
Step 4: Expand the sine term in the numerator using the difference formula. Recall the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Let $A = (x-a)$ and $B = (x-b)$. \[ I = \frac{1}{\sin(b-a)} \int \frac{\sin(x-a)\cos(x-b) - \cos(x-a)\sin(x-b)}{\sin(x-a) \sin(x-b)} dx \]
Step 5: Split the fraction into two terms. \[ I = \frac{1}{\sin(b-a)} \int \left( \frac{\sin(x-a)\cos(x-b)}{\sin(x-a) \sin(x-b)} - \frac{\cos(x-a)\sin(x-b)}{\sin(x-a) \sin(x-b)} \right) dx \] Simplify each term: \[ I = \frac{1}{\sin(b-a)} \int \left( \frac{\cos(x-b)}{\sin(x-b)} - \frac{\cos(x-a)}{\sin(x-a)} \right) dx \] \[ I = \frac{1}{\sin(b-a)} \int \left( \cot(x-b) - \cot(x-a) \right) dx \]
Step 6: Integrate the cotangent terms. Recall that $\int \cot u \, du = \log|\sin u| + C$. \[ I = \frac{1}{\sin(b-a)} \left( \int \cot(x-b) \, dx - \int \cot(x-a) \, dx \right) \] \[ I = \frac{1}{\sin(b-a)} \left( \log|\sin(x-b)| - \log|\sin(x-a)| \right) + C \]
Step 7: Combine the logarithmic terms. Using the logarithm property $\log M - \log N = \log \frac{M}{N}$: \[ I = \frac{1}{\sin(b-a)} \log \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C \] where $C$ is the constant of integration.