1. ## integration

How would you take the integral of sec^2(3x)?
The ^2 thing messes me up.

2. $\displaystyle \displaystyle \int\sec ^23xdx=\int\frac{dx}{\cos ^23x}=\frac{1}{3}\int \frac{3}{\cos ^23x}dx=\frac{1}{3}\tan 3x+C$.

Or by substitution $\displaystyle 3x=t\Rightarrow 3dx=dt\Rightarrow dx=\frac{1}{3}dt$
Then $\displaystyle \frac{1}{3}\int\frac{dt}{\cos ^2t}=\frac{1}{3}\tan t+C$.
Back substitution $\displaystyle \frac{1}{3}\tan 3x+C$

3. Originally Posted by red_dog
$\displaystyle \displaystyle \int\sec ^23xdx=\int\frac{dx}{\cos ^23x}=\frac{1}{3}\int \frac{3}{\cos ^23x}dx=\frac{1}{3}\tan 3x+C$.

Or by substitution $\displaystyle 3x=t\Rightarrow 3dx=dt\Rightarrow dx=\frac{1}{3}dt$
Then $\displaystyle \frac{1}{3}\int\frac{dt}{\cos ^2t}=\frac{1}{3}\tan t+C$.
Back substitution $\displaystyle \frac{1}{3}\tan 3x+C$
how did you get from 3/cos^2(3x) = tan3x?
tan is sin/cos, but there isnt a sin

4. $\displaystyle (\tan x)'=\frac{1}{\cos ^2x}\Rightarrow \int\frac{1}{\cos ^2x}dx=\tan x+C$
$\displaystyle (\tan u)'=\frac{u'}{\cos ^2u}\Rightarrow \int\frac{u'}{\cos ^2u}du=\tan u+C$