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How do I evaluate #intsec^2(x) tan^2(x) dx#?

1 Answer

Jul 13, 2018

The answer is #=(tan^3x)/3+C#

Explanation:

Perform this integral by substitution

Let #u=tanx#, #=>#, #du=sec^2xdx#

The integral is

#I=intsec^2xtan^2xdx#

#=intu^2du#

#=u^3/3#

#=(tan^3x)/3+C#

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