- #1

Nenad

- 698

- 0

[tex] \int \sqrt{x^2 - 1} dx [/tex]

so far:

[tex] x = sec \theta [/tex]

[tex] \frac{dx}{d \theta} = sec \theta tan \theta [/tex]

[tex] dx = sec \theta tan \theta d\theta [/tex]

now we substitute:

[tex] \int \sqrt{x^2 - 1} dx [/tex]

[tex]= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta [/tex]

since [tex] sec^2 \theta - 1 = tan^2 \theta [/tex]

[tex]= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta = \int \sqrt{tan^2 \theta} sec \theta tan \theta d \theta [/tex]

[tex]= \int tan^2 \theta sec \theta d \theta [/tex]

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

Regards,

Nenad