I am trying to do this:
integral sign xsec^2(3x)dx
I am stuck on how to do it help plz. How do you do these types of problems?
Hello, davecs77!
$\displaystyle \int x\sec^2(3x)\,dx$
By parts: .$\displaystyle \begin{array}{ccccccc}u & = & x & \quad & dv & = & \sec^2(3x)\,dx \\ du & = & dx & \quad & v & = & \frac{1}{3}\tan(3x)\end{array}$
Then: .$\displaystyle \frac{1}{3}x\tan(3x) - \frac{1}{3}\int\tan(3x)\,dx \;=\;\frac{1}{3}x\tan(3x) + \frac{1}{9}\ln\left|\cos(3x)\right| + C$