Question

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

Answer 1

We are given that $I=\int \frac{\cos (x-a)}{\sin (x+b)}dx$

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

$=\int \frac{\cos \left[\right(x+b)-(a+b\left)\right]}{\sin (x+b)}dx$

$=\int \frac{\left[\cos \right(x+b\left)\cos \right(a+b)+\sin (a+b)+\sin (x+b\left)\sin \right(a+b\left)\right]}{\sin (x+b)}dx$

$=\int \cot (x+b)\cos (a+b)dx+\int \sin (a+b)dx$

as we know, $\int \cot \theta =\ln |\sin \theta |+C$

$I=\cos (a+b).\ln |\sin (x+b)|+\left[\sin \right(a+b\left)\right]x+C$