How do you calculate the anti-derivative of #secx(secx + tanx)dx#?

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Answer 1 · 2015-02-14T14:05:44

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

This integral becomes:

#intsecx(secx+tanx)dx=int1/cosx(1/cosx+sinx/cosx)dx=#

#=int(1/cos^2x+sinx/cos^2x)dx=int(1/cos^2x-(-sinx)cos^-2x)dx=#

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

Remembering that:

#int[f(x)]^nf'(x)dx=[f(x)]^(n+1)+c#.