Trigonometric Integrals problem, more help

[afcwestwarrior]

1. The problem statement, all variables and given/known data

∫sec^6 t

2. Relevant equations
sec^2 (x)=1+tan^2 (x)


3. The attempt at a solution
∫ (sec^2 (t))^2 * (Sec^2 (t))^2

∫(1+tan^2 (x))^2 (1+tan^2 (x))

∫(1+2tan^2 (x) +tan^4 (x)) (1+tan^2 (x))
u=tan (x) du=sec^2 dx

i'm stuck now

[rock.freak667]

1. The problem statement, all variables and given/known data

∫sec^6 t dt



Always remeber to put what you integrating with respect to.

I'll help you out a bit

\int sec^6(t) dt= \int sec^4(t)*sec^2(t) dt

Now use sec2=1+tan2 to find what sec4will be.