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

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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#