How do you integrate #(1-(tanx)^2)/(secx)^2#?

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Answer 1 · 2016-12-28T22:46:14

I got #1/2sin(2x) + C#.

Note the identity:

#1 + tan^2x = sec^2x#,

so that:

#int (1 - tan^2x)/(sec^2x)dx#

#int (2 - sec^2x)/(sec^2x)dx#

#int 2cos^2x - 1dx#

Now, note another identity:

#cos^2x = (1 + cos2x)/2#

Now this is much simpler to integrate than we started with!

#=> int 2((1 + cos2x)/2) - 1dx#

#= int cos2xdx#

Therefore:

#=> color(blue)(int (1 - tan^2x)/(sec^2x)dx = 1/2sin2x + C)#