Question

Evaluate
$I=\int \frac{1}{\sin (x-a)\cos (x-b)}dx$

Answer 1

We have,

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

On multiplying and divide by $\cos (b-a)$

Then,

$I=\frac{1}{\cos (b-a)}\int \frac{\cos [(x-a)-(x-b)]}{sn(x-a)\cos (x-b)}dx$

$I=\frac{1}{\cos (b-a)}\int \frac{\cos (x-a)\cos (x-b)+\sin (x-a)\sin (x-b)}{\sin (x-a)\cos (x-b)}dx$

$I=\frac{1}{\cos (b-a)}(\int \frac{\cos (x-a)}{\sin (x-a)}dx+\int \frac{\sin (x-b)}{\cos (x-b)}dx)$

$I=\frac{1}{\cos (b-a)}(\int \cot (x-a)dx+\tan (x-b)dx)$

$I=\frac{1}{\cos (b-a)}[\log \left|\sin (x-a)\right|+\log \left|\sec (x-b)\right|]+C$

$I=\frac{1}{\cos (b-a)}\left[\log \left|\frac{\sin (x-a)}{\cos (x-b)}\right|\right]+C$

Hence, this is the answer.