How do you find the integral of #sec^5 x dx#?

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Aug 29, 2017

Answer:

#intsec^5xdx=1/4sec^3x*tanx+3/8secxtanx+3/8lnabs(secx+tanx)+C#

Explanation:

at first we will break #intsec^5xdx # into #intsec^3x*sec^2xdx# and by using #intudv=uv-intvdu# (integration by parts) we will take #u=sec^3x# and #dv=sec^2xdx#.

These show that #du=3sec^2cx(secxtanx)=3sec^3xtanx# and #v=tanx#.

Then:

#intsec^3*sec^2xdx#
#=sec^3xtanx-inttanx*3sec^3xtanxdx#

Grouping #tan^2x# and rewriting as #sec^2x-1#:

#=sec^3xtanx- int3sec^3x(sec^2x-1)dx#
#=sec^3xtanx-int(3sec^5x-3sec^3x)dx#

So

#intsec^5xdx=sec^3x*tanx-3intsec^5xdx+3intsec^3xdx#
#4intsec^5xdx=sec^3x*tanx+3intsec^3xdx#

And we know that#intsec^3xdx=1/2secxtanx+1/2lnabs(secx+tanx)#

So

#4intsec^5xdx=sec^3x*tanx+3(1/2secxtanx+1/2lnabs(secx+tanx))#

#intsec^5xdx=1/4sec^3x*tanx+3/8secxtanx+3/8lnabs(secx+tanx)+C#

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