How do you find the integral of #tan^2 (x)sec(x) dx#?

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Answer 1 · 2015-04-25T16:09:10

#tan^3x /3# +C

This can be easily be solved by using substitution. Let tan x =u, so that #sec^2x# dx=du

#int tan^2x sec x dx#= #intu^2du#= #u^3/3# +C=#tan^3x /3# +C

Answer 2 · 2015-04-26T06:56:22

#1/2 secx tanx -1/2 ln(secx +tanx)#

Rewrite integral as, #int(sec^2x-1)sec xdx#= #intsec^3xdx-int secx dx#. Now solve #int sec^3x dx# by integrating by parts, #(sec x) sec^2 x#

=sec x tanx - #int secx tanx tanx dx# = sec x tan x -#int secx tan^2 xdx#

= secx tanx -#int secx (sec^2x-1)dx# = secx tanx - #int sec^3x dx + int sec xdx#. Now transpose #sec^3xdx# to the other side and have

#intsec^3x dx#= #1/2 secx tan x +1/2 int secx dx#. Thus it is,

#int tan^2x secx dx= 1/2 secx tan x +1/2 int secx dx -int sec xdx#

= #1/2 sec x tan x -1/2int secx dx#

= #1/2 secx tan x-1/2 ln(secx+tanx)# +C