Find the integral `intsec(x)/(sec(x)+tan(x))dx`?

`I=intsecx/(secx+tanx)dx,`

`=intsecx/(secx+tanx)xx(secx-tanx)/(secx-tanx)dx,`

`=int(sec^2x-secxtanx)/(sec^2x-tan^2x)dx,`

`=intsec^2xdx-intsecxtanxdx,...[because, sec^2x-tan^2x=1],`

`:. I=tanx-secx+C.`

`intsecx/(secx+tanx)dx`

Let `u=secx+tanx`, then

`du=(secxtanx+sec^2x)dx=secx(secx+tanx)dx`

`intsecx/(secx+tanx)dx`

= `intsecx/uxx(du)/(secx(secx+tanx)`

= `int1/u^2du`

= `-1/u +C`

= `-1/(secx+tanx)+C`

Given: `intsec(x)/(sec(x)+tan(x))dx`

Multiply by 1 in the form of `(sec(x)-tan(x))/(sec(x)-tan(x))`

`intsec(x)/(sec(x)+tan(x))(sec(x)-tan(x))/(sec(x)-tan(x))dx`

The denominator the difference of two squares:

`int(sec(x)(sec(x)-tan(x)))/(sec^2(x)-tan^2(x))dx`

From the identity `1 + tan^2(x)=sec^2(x)`, we see that the denominator becomes 1:

`intsec(x)(sec(x)-tan(x))dx`

Distributing the secant function gives us two integrals:

`intsec^2(x)dx-intsec(x)tan(x)dx`

The first integral becomes the tangent function:

`tan(x)-intsec(x)tan(x)dx`

Use the identities `tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)` on the second integral:

`tan(x)-intsin(x)/cos^2(x)dx`

let `u = cos(x)`, then `du = -sin(x)dx`

`tan(x)+intu^-2du`

`tan(x)-u^-1+C`

`intsec(x)/(sec(x)+tan(x))dx= tan(x) -sec(x)+ C`