Integrating tan(3x) using substitution method?

[tandoorichicken]

How do you integrate tan(3x)?
As in
$$\int \tan{3x} \,dx$$

 

[tandoorichicken]

nevermind i got it

 

[M98Ranger]

Integrating tan(3x) can be done using substitution, specifically the substitution u = 3x. This will result in the integral becoming \int \frac{\tan u}{3} \,du. We can then use the trigonometric identity \tan x = \frac{\sin x}{\cos x} to rewrite the integral as \frac{1}{3} \int \frac{\sin u}{\cos u} \,du. From here, we can use the substitution v = \cos u, which will result in the integral becoming \frac{1}{3} \int \frac{1}{v} \,dv. This can be easily evaluated to be \frac{1}{3} \ln|v| + C. Substituting back in for u and then x, we get the final result of \frac{1}{3} \ln|\cos 3x| + C.