Question

What is `int x/(1+sinx) dx`?

Answer 1

`x(tanx-secx)+ln|(secx+tanx)/secx|+c`

Explanation:

`I=intx/(1+sinx)dx=int(x(1-sinx))/(1-sin^2x)dx=int(x(1-sinx))/cos^2xdx`
`=>I=intxsec^2xdx-intxsecxtanxdx`
Integration by parts,we get
`I=[xtanx-int1*tanxdx]-[x*secx-int1*secxdx]`
`I=xtanx-ln|secx|-xsecx+ln|secx+tanx|+c`
`=xtanx-xsecx+ln|secx+tanx|-ln|secx|+c`
`=x(tanx-secx)+ln|(secx+tanx)/secx|+c`