Question

How do Integrate ∫ sen (3x +1)/1+cos (3x +1)?

Answer 1

`(-1/3)ln(cos(3x+1))+k`

Explanation:

considering sen as sin
let `1+cos(3x+1)=t`
`rArr -3sin(3x+1)dx = dt`
`rArr sin(3x+1)dx = (-1/3)dt`
so given integral become
`int (-1/3)dt/t`

`rArr (-1/3)lnt +k`
substituting `t` back
`(-1/3)ln(cos(3x+1))+k`

more simplified version would be
take constant `k` as `lnk`
`(-1/3)ln(k*cos(3x+1))`