Can you find this integral?

Question

∫ (e^(t/2))(2cos(t))dt.

Answer 1

Use integration by parts twice: ∫ u dv = uv - ∫ v du.

First iteration: u = 2e^t/2, du = e^t/2 dt, dv = cos t dt, v = sin t.

I = ∫ 2e^t/2 cos t dt = 2e^t/2 sin t - ∫ e^t/2 sin t dt = 2e^t/2 - J.

Second iteration: u = e^t/2, du = (1/2)e^t/2 dt, dv = sin t dt, v = -cos t.

J = ∫ e^t/2 sin t dt = -e^t/2 cos t - ∫ (-1/2)e^t/2 cos t dt = -e^t/2 cos t + I/4.

Substitute and solve

I = 2e^t/2 sin t + e^t/2 cos t - I/4

5I/4 = e^t/2(2sin t + cos t)

I = (4/5)e^t/2(2sin t + cos t)

Hence ∫ 2e^t/2 cos t dt = (4/5)e^t/2(2sin t + cos t) + C