Question

How do you integrate `int sec^3(pix)`?

Answer 1

`I=1/pi[tan(pix)/2sec(pix)+1/2ln|tan(pix)+sec(pix)|]+c`

Explanation:

Here,

`I=intsec^3(pix)dx`

`=intsec(pix)sec^2(pix)dx`

`I=intsqrt(1+tan^2(pix))sec^2(pix)dx`

Let,`tan(pix)=u=>sec^2(pix)*pidx=du`

`=>sec^2(pix)dx=1/pidu`

So,

`I=1/piintsqrt(1+u^2)du`

`=1/pi[u/2sqrt(1+u^2)+1/2ln|u+sqrt(1+u^2)|]+c`

Subst.`,u=tan(pix)`

#=1/pi[tan(pix)/2sqrt(1+tan^2(pix))+1/2ln|tan(pix)+sqrt(1+tan^2(pi x))|]+c#

`I=1/pi[tan(pix)/2sec(pix)+1/2ln|tan(pix)+sec(pix)|]+c`