Question

$\displaystyle \int \frac{dx}{\sin(x-a)\cos(x-b)} =$

• A. $\dfrac{1}{\sin(a - b)} \log\left| \dfrac{\sin(x - a)}{\cos(x - b)} \right| + C$
• B. $\dfrac{1}{\cos(b - a)} \log\left| \dfrac{\sin(x - a)}{\cos(x - b)} \right| + C$
• C. $\dfrac{1}{\cos(b - a)} \log\left| \sin(x - a)\cos(x - b) \right| + C$
• D. $\dfrac{1}{\sin(a - b)} \log\left| \sin(x - a)\cos(x - b) \right| + C$

Hint:

Trick:

• Use identity to split product into sum of functions
• Then integrate standard forms: $\int \cot u\, du$, $\int \tan u\, du$

Answer 1

The Correct Option is B

Let $I = \int \frac{dx}{\sin(x-a)\cos(x-b)}$ Multiply numerator and denominator by $\cos(b - a)$: \[ I = \frac{1}{\cos(b - a)} \int \frac{\cos(b - a)}{\sin(x - a)\cos(x - b)} dx \] Use identity: \[ \cos(b - a) = \cos(x - a)\cos(x - b) + \sin(x - a)\sin(x - b) \] So: \[ I = \frac{1}{\cos(b - a)} \int \left( \cot(x - a) + \tan(x - b) \right) dx \] Integrate: \[ = \frac{1}{\cos(b - a)} \left[ \log|\sin(x - a)| - \log|\cos(x - b)| \right] + C \Rightarrow \boxed{\frac{1}{\cos(b - a)} \log\left| \frac{\sin(x - a)}{\cos(x - b)} \right| + C} \]