What is the antiderivative of #tanx dx#?

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Answer 1 · 2015-10-07T21:57:16

The antiderivative of #tan(x)# is #-ln(cos(x))#

We know #tan(x) = sin(x)/cos(x)#.

#inttan(x)dx = intsin(x)/cos(x)dx#

We can use substitution to simplify the integral.

#u = cos(x)# and #du = -sin(x)dx#

#inttan(x)dx = intsin(x)/cos(x)dx = -int1/udu = -ln(u) + C#

To get the answer, just plug #u# back in to get:

#-ln(cos(x))#