Question

Evaluate:

`int sin(x + a)/cos(x - b)dx`

Answer 1

`int sin(x + a)/cos(x - b)dx`

= `intsin[(x - b) + (a + b)]/cos(x - b)dx`

= `int(sin(x - b) * cos(a + b) + cos(x - b) * sin(a + b))/(cos(x - b))dx`

= `int[(sin(x - b) * cos(a + b))/cos(x - b) + (cos(x - b) * sin(a + b))/(cos(x - b)]]dx` 

= `int[tan(x - b) * cos(a + b) + sin(a + b)]dx`

= `cos(a + b) inttan(x - b)dx + sin(a + b) int 1 dx`

= cos (a + b) log|sec (x − b)| + [sin (a + b)]x + c

= cos (a + b) log|sec (x − b)| + x sin (a + b) + c