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

Question

5310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-07T06:11:32

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

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. $,u=tan(pix)$

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

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

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