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