Question

Question #e4efb

Answer 1

`int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C`

Explanation:

Making `y = x+a` and `dy = dx` we have

`int sinx/sin(x+a) dx = int sin(y-a)/siny dy = int (cosa - cos y/siny sina)dy = y cosa-log(siny) sina + C`

and finally

`int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C`