How do you integrate #int tan 3x+sec3x dx#?

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Answer 1 · 2016-07-15T01:45:03

#int tan3x + sec3x dx= 1/3 ln|sec3x + tan3x| - 1/3 ln|cos3x| + C#

This integral can be solved using a u-substitution, provided that

#int tan u du = -ln|cos u| + C# and

#int sec u du = ln|sec u + tan u| + C#

In our integral, we can let #u = 3x -> du = 3 dx -> 1/3 du = dx#

This way, we can separate these two expression by the properties of integrals, so our integral becomes

#int tan3x + sec3x dx = 1/3 int tanu du + 1/3 int secu du #

#= -1/3 ln|cos3x| + 1/3 ln|sec3x + tan3x| + C#

#= 1/3 ln|sec3x + tan3x| - 1/3 ln|cos3x| + C#