# Tough Calculus Integral problem

• March 21st 2010, 07:57 PM
dtewfik
Tough Calculus Integral problem
Looking to solve sec^3(x)dx
and I can't figure it out. Any help guys? Thank you so much,
• March 21st 2010, 09:22 PM
pickslides
Does this help?

$\int\sec^3x~dx = \int\sec^2x \sec x~dx = \int\frac{1+\tan^2 x}{ \cos x}~dx = \int\frac{1}{ \cos x}~dx+ \int\frac{\tan^2 x}{ \cos x}~dx$

Remember $\tan x = \frac{\sin x }{\cos x}$ and $\frac{d}{dx}(\sin x) = \cos x$