Simple Integral w/ Trigonometric Substitution

1. May 9, 2005

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,

2. May 9, 2005

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.

3. May 9, 2005

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