Question

What is the antiderivative of `1/ (1+sinx) dx`?

Answer 1

`I=tanx-secx+c`

Explanation:

We know that,
`color(red)((1)sin^2theta+ cos^2theta=1`
`color(blue)((2)1/costheta=sectheta and sintheta/costheta=tantheta`
`color(violet)((3)intsec^2xdx=tanx+c`
`color(violet)((4)intsecxtanxdx=secx+c`
Here,

`I=int1/(1+sinx)dx=int((1-sinx))/((1+sinx)(1-sinx))dx`

`=>I=int(1-sinx)/(1-sin^2x)dx`

`=int(1-sinx)/cos^2xdx...tocolor(red)(Apply(1)`

`=int[1/cos^2x-sinx/cos^2x]dx`

`=int[sec^2x-secxtanx]dx...tocolor(blue)(Apply(2)`

Using , `color(violet)((3) and (4)`, we get

`I=tanx-secx+c`