can someone help me find the integral of sec cubed/tan?
I would really appreciate it!
$\displaystyle \int\frac{sec^3\theta}{tan \theta}d\theta$
You can convert secant and tangent into sines and cosines.
$\displaystyle \int\frac{\frac{1}{cos^3\theta}}{\frac{sin\theta}{ cos\theta}}d\theta$
$\displaystyle =\int\frac{cos\theta}{sin\theta * cos^3\theta}d\theta$
$\displaystyle =\int\frac{1}{sin\theta*cos^2\theta}d\theta$
$\displaystyle =\int sec^2\theta*csc\theta d\theta$
can you take it from there, you do it by parts.