# Trig Integral

1. Feb 5, 2009

### Shambles

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

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations
u = 1+tant
du = sec^2(t) dt
dt = du / sec^2(t)

3. The attempt at a solution

It seems like I should be using substitution in the equation, however the exponent is messing things up for me. I recall from derivatives how they interact with the chain rule, but am not sure how this would work backwards in integration. Something like,

I(u^3)(sec^2(t)) = (u^4/4)((sec^2(t)) (tan(t))

Except I haven't gotten rid of the t variable and now have t and u. Any points are welcome.

2. Feb 5, 2009

### Hitman2-2

Why don't you just substitute u for (1 + tan t) and du for sec^2(t) dt (and take care of the limits of integration, of course)?

3. Feb 5, 2009

### Shambles

Ah I see how when I change the limits of integration it removes the nasty sec^2(t) so all i'm left with is the integral of u^3 with u going from 1 to 2. Thanks.