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

1 Answer
Jul 15, 2016

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

Explanation:

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