If \(\sin(a-b) \ne 0\), then \(\int\displaystyle\frac{dx}{\cos(x-a)\cos(x-b)}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle \mathrm{cosec}(a-b)\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C\)
Since \(a-b = (x-b)-(x-a)\), the sine of a difference gives
\( \sin(a-b) = \sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a) \)
Dividing by \(\cos(x-a)\cos(x-b)\):
\( \displaystyle \frac{\sin(a-b)}{\cos(x-a)\cos(x-b)} = \tan(x-b)-\tan(x-a) \)
So the integrand is
\( \displaystyle \frac{1}{\cos(x-a)\cos(x-b)} = \frac{1}{\sin(a-b)}\left[\tan(x-b)-\tan(x-a)\right] \)
Using \(\int\tan(x-c)\,dx = -\log\left|\cos(x-c)\right|\):
\( \displaystyle I = \frac{1}{\sin(a-b)}\left[-\log\left|\cos(x-b)\right|+\log\left|\cos(x-a)\right|\right]+C \)
\( \displaystyle = \mathrm{cosec}(a-b)\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C \)
Why the other options are incorrect
Option (1) \(\sin(a-b)\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C\) — Incorrect. The integrand was multiplied by \(\sin(a-b)\), so the result must be divided by it, not multiplied.
Option (2) \(\mathrm{cosec}(a-b)\log\left|\frac{\cos(x-b)}{\cos(x-a)}\right|+C\) — Incorrect. The fraction inside the log is upside down, which reverses the sign of the whole answer.
Option (4) \(\sec(a-b)\log\left|\frac{\cos(x-a)}{\cos(x-b)}\right|+C\) — Incorrect. The identity used is for \(\sin(a-b)\), so the factor is \(\mathrm{cosec}(a-b)\), not \(\sec(a-b)\).