Integrate #intsec^3(x)# by using integral by parts??

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Answer 1 · 2018-05-01T19:30:00

#I=1/2secxtanx+1/2ln|secx+tanx|+C#

Here,

#color(red)(I=intsec^3xdx...to(1)#

#=intsecxsec^2xdx#

#"Using "color(blue)"Integration by Parts"#

#int(u*v)dx=uintvdx-int(u'intvdx)dx#

Let. #u=secx and v=sec^2x#

#=>u'=secxtanx and intvdx=tanx#

#I=secx xxtanx-intsecxtanx xxtanx dx#

#=secxtanx-intsecxtan^2xdx+c#

#=secxtanx-intsecx(sec^2x-1)dx+c#

#I=secxtanx-color(red)(intsec^3xdx)+intsecxdx+c#

#I=secxtanx-color(red)(I)+intsecxdx+c...to #[ from #(1) #]

#I+I=secxtanx+intsecxdx+c#

#2I=secxtanx+ln|secx+tanx|+C#

#I=1/2secxtanx+1/2ln|secx+tanx|+C#