Question

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

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

Answer 1

The correct option is B $\frac{1}{\cos (a-b)}\log \left|\frac{\sin (x-a)}{\cos (x-b)}\right|+C$

$I=\int \frac{dx}{\sin (x-a)\cos (x-b)}$
Multiply numerator and denominator by a constant $cos(a-b)$
$=\frac{1}{\cos (a-b)}\int \frac{\cos (a-b)}{sin(x-a)cos(x-b)}dx$
We know $\cos (a-b)=\cos ((x-b)-(x-a))=\cos (x-b)\cos (x-a)+\sin (x-b)\sin (x-a)$
$=\frac{1}{\cos (a-b)}\int \frac{\cos (x-b)\cos (x-a)+\sin (x-b)\sin (x-a)}{sin(x-a)cos(x-b)}dx$
$=\frac{1}{\cos (a-b)}\int \{\frac{\cos (x-b)\cos (x-a)}{sin(x-a)cos(x-b)}dx+\frac{\sin (x-b)\sin (x-a)}{sin(x-a)cos(x-b)}dx\}$
$=\frac{1}{\cos (a-b)}\{\int \frac{\cos (x-a)}{sin(x-a)}dx+\int \frac{\sin (x-b)}{cos(x-b)}dx\}$
$=\frac{1}{\cos (a-b)}\left\{\log \left(\sin (x-a)\right)-\log \left(\cos (x-b)\right)\right\}+C$
$=\frac{1}{\cos (a-b)}\log \left|\frac{\sin (x-a)}{cos(x-b)}\right|+C$