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How do I evaluate #int(sec^2(x))/(tan^2(x)) d x#?

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Mar 4, 2015

The answer is: #-1/tanx+c#.

#intsec^2x/tan^2xdx=int1/cos^2x*tan^(-2)xdx=#

#=tan^(-2+1)x/(-2+1)+c=-1/tanx+c#.

This is because:

#int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c# and the derivative of the function #tanx# is #1/cos^2x.#

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