How do you find the antiderivative of # 1/(cos x)^2#?

1 Answer
Morgan
Nov 12, 2016

Use #sec(x)=1/(cosx)# and rewrite as #intsec^2(x)#, where the anti-derivative of #sec^2x = tanx#.

Explanation:

This can also be written as:

#int1/(cosx)^2dx#

This one is a bit tricky in that you have to recognize that #1/cosx^2# is equivalent to #sec^2x#, but once you've figured that out, it's quite simple.

#=>intsec^2(x)dx#

The anti-derivative of #sec^2x# is simply #tan(x)#. We also account for any constants that could be present by adding a #+C# at the end of our answer. Thus,

#int1/(cosx)^2dx=intsec^2(x)dx = tan(x)+C#

Related questions
See all questions in Integrals of Trigonometric Functions
Impact of this question
3056 views around the world
You can reuse this answer
Creative Commons License