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:

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

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