How do you integrate #int sec^2xtanx#?

2 Answers
Andrea S.
Nov 24, 2016

You have to remeber that #sec^2(x)# is the derivative of #tan(x)#

Explanation:

So:

#int sec^2 x tan x dx = int tan x d (tan x) = (tan^2 x ) / 2#

mason m
Nov 24, 2016

#sec^2x/2+C# or #tan^2x/2+C#

Explanation:

You can also see this this way:

#intsec^2xtanxdx=intsecx(secxtanx)dx#

If #u=secx# then #du=secxtanxdx# so

#=intudu=u^2/2=sec^2x/2+C#

This is equivalent to the other answer of #tan^2x/2+C# because #tan^2x# and #sec^2x# are only a constant away from one another through the equality #tan^2x+1=sec^2x#.

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