How do you find the antiderivative of $int sin(tanx)/cos^2x dx$ ?

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Answer 1 · 2016-11-22T14:08:58

The derivative of $tanx$ is $sec^2 x$ which is $1/cos^2x$ . Use substitution.

$int sin(tanx)/cos^2x dx = int sin(tanx) (sec^2x) dx$

With $u = tanx$ , this is $int sinu du = -cosu +C$ .

So we get

$int sin(tanx)/cos^2x dx = -cos(tanx) +C$

(Check the answer by differentiation.)

How do you find the antiderivative of int sin(tanx)/cos^2x dxint sin(tanx)/cos^2x dx?