Looking to solve sec^3(x)dx

and I can't figure it out. Any help guys? Thank you so much,

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- Mar 21st 2010, 07:57 PMdtewfikTough Calculus Integral problem
Looking to solve sec^3(x)dx

and I can't figure it out. Any help guys? Thank you so much, - Mar 21st 2010, 09:22 PMpickslides
Does this help?

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

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