Question

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

Answer 1

$I=\int \frac{cos(x-a)}{sin(x+b)}dx$
Put
x + b = t
$\Rightarrow dx=dt$
$=\int \frac{cos(t-b-a)}{sint}dt$
$=\int \frac{cos\{t-(a+b\left)\right\}}{sint}dt$
$=\int \frac{costcos(a+b)+sintsin(a+b)}{sint}dt$
$=\int \left[cottcos\right(a+b)+sin(a+b\left)\right]dt$
$=cos(a+b)\int cottdt+sin(a+b)dt$
$=cos(a+b)In\int sint1+tsin(a+b)sin(a+b)+c$
$=cos(a+b)In\int sin(x+b)1+(x+b)sin(a=b)+c$
$=cos(a+b)In|sin(x+b)|+xsin(a+b)+c$
$c=c+bsin(a+b)$