Integration problem involving secant and tangent

[Lizzy]

I need help integrating this problem:

the integral of (sec t)^3(tan t)^3 dt

Would you need to set it up as the integral of

(sec t)^2[(sec t)^2 - (1)](sec t tan t)dt

Thanks for the help!!

 

[mark07]

Exactly...

When you do that, your integral becomes

\(\displaystyle \L \int (\sec t)^2\left[ (\sec t)^2 - 1 \right] (\sec t \tan t) dt\)

and letting $\displaystyle u = \sec t$ gives you

\(\displaystyle \L \int u^2 \left[ u^2 - 1 \right] du\)

and I'm sure you can integrate that, right?