Question

If $c$ is an arbitrary constant then $\int \frac{\cos (x+a)}{\sin (x+b)}dx=$

• A. $\cos (a-b)\ln \left|\sin (x-b)\right|-x\sin (a-b)+c$
• B. $\cos (a-b)\ln \left|\sin (x+b)\right|-x\sin (a-b)+c$
• C. $\cos (a+b)\ln \left|\sin (x+b)\right|-x\sin (a+b)+c$
• D. $\cos (a-b)\ln \sin \left|(x+b)\right|-x\sin (a+b)+c$
• E. $\cos (a-b)\ln \left|\sin (x+b)\right|+x\sin (a-b)+c$

Answer 1

The correct option is B $\cos (a-b)\ln \left|\sin (x+b)\right|-x\sin (a-b)+c$

$\int \frac{\cos (x+a)}{\sin (x+b)}dx$
$=\int \frac{\cos (x+b+a-b)}{\sin (x+b)}dx$
$=\int \frac{\cos (x+b)\cos (a-b)-\sin (x+b)\sin (a-b)}{\sin (x+b)}dx$
$=\cos (a-b)\int \cot (x+b)dx-\sin (a-b)\int dx$
$=\cos (a-b)\ln |\sin (x+b)|-x\sin (a-b)+C$