Simple Integral w/ Trigonometric Substitution

[Nenad]

Hello everyone, Im having some trouble with an integral.
\int \sqrt{x^2 - 1} dx

so far:
x = sec \theta
\frac{dx}{d \theta} = sec \theta tan \theta
dx = sec \theta tan \theta d\theta

now we substitute:

\int \sqrt{x^2 - 1} dx
= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta

since sec^2 \theta - 1 = tan^2 \theta

= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta = \int \sqrt{tan^2 \theta} sec \theta tan \theta d \theta

= \int tan^2 \theta sec \theta d \theta

this is where Im stuck. A hint would be appreciated. Thanks in advance

Regards,

Nenad

[whozum]

Express tangent in terms of secant, separate the two integrals. One will be trivial, and If i remember correctly the other can be done with a substitution.

[Nenad]

Express tangent in terms of secant, separate the two integrals. One will be trivial, and If i remember correctly the other can be done with a substitution.

thanks, I got it, but I had to use Integration by parts like 5 times to get \int sec^3 \theta d\theta

Regards,

Nenad