Integration problem...?

Question

Integrate sqrt( e^(2t) + 2 + e^(-2t)) dt from t = 0 to t = ln(3).

Thanks.

Answer 1

sqrt( e^(2t) + 2 + e^(-2t))
= sqrt [ ( e^t + e^(-t) }^2
= e^t + e^(-t)

Integral [ e^t + e^(-t) ] dt
= e^t - e^(-t) + C

Putting boundary conditions,
= e^(ln3) - e^(-ln3) - e^0 + e^0
= 3 - 1/3
= 8/3

Answer 2

sqrt( e^(2t) + 2 + e^(-2t))

Let u = e^(t)

sqrt(u^2 + 2 + 1/u^2)
= (u + 1/u)
= e^t + e^-t

Integrate this to get:
e^t - e^-t
Apply the limits to get
[e^ln(3) - e^-ln(3)] - [e^0 - e^0]
= 3 - 1/3
= 8/3

Answer 3

Your integrand has (e^t + e^(-t))^2 under the radical, so this is just integ(e^t + e^(-t))dt = e^t - e^(-t) = (3 - 1/3) - (1 - 1) = 8/3.