Calculate the following integral:

Question

∫ sec^4 (3x)/ tan^3 (3x) dx

For this one, can I bring up the tan to tan^-3?

Answer 1

yes, so you have

∫sec^4(3x)* tan^-3(3x) dx
∫sec^2(3x)*(tan^2(3x)+1)*tan^-3(3x) dx

Now let u = tan(3x)
du = 3sec^2(3x) dx

∫1/u + 1/u^3 du/3
= 1/3 ln(u) - 1/6 u^-2
= 1/3 ln(tan(3x) - 1/6 cot^2(3x) + C

Answer 2

thanks Steve! You're a lifesaver