I can't for the life of me evaluate this integral using the substitution method, ∫(sec3x)^2 dx?
The indefinite integral is ∫(sec3x)^2 dx
So I am taught that we turn ∫f(g(x))(g'x)dx into ∫f(u)du. Practicing this, I turned
∫(sec3x)^2 dx into ∫f(u) du, where u = sec(3x), and du = 3sec(3x)tan(3x)dx. Then sec(3x)tan(3x)dx = 3du. But there is no g'(x) or sec(3x)tan(3x) in the integrand for me to get to the next step. So unless I am suppose to put it outside of the integrand along with the 3, and I have tried that, I don't know what to do.
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Verified answer
Let u = 3x
du = 3
(1/3) ∫ (sec(u))^2
= (1/3)(tan(u)) + C
= (1/3)(tan(3x) + C
Take derivative to double check:
(1/3)(sec^2(3x)) * 3
= sec^2 (3x)