Question

How do you find the antiderivative of `sin(pix) dx`?

Answer 1

`" "intsin(pix)dx=-1/picos(pix)+C`

Explanation:

Let
`" "`
`u=cos(pix)`
`" "`
`du=-pisin(pix)dx`
`" "`
`-(du)/pi=sin(pix)dx`
`" "`
`intsin(pix)dx=int-(du)/pi=-1/piintdu=-1/pi(u)+C`
`" "`
Therefore,`" "intsin(pix)dx=-1/picos(pix)+C`