Question

Integral help. Please

Answer 1

\[\int_0^\pi ie^{2it/3}dt\]

Answer 2

@freckles

Answer 3

assuming i is a constant and equal to sqrt(-1), then let u=2it/3;
du=2idt/3, idt=3du/2 and
t=0, u=0 and t=pi, u=2ipi/3

the integral becomes

\[\int\limits_{0}^{\pi}ie ^{\frac{ 2i t }{ 3 }}dt = \int\limits_{0}^{\frac{ 2i \pi }{ 3 }}e ^{u}\frac{ 3 }{ 2}du\]

Answer 4

use \[e ^{i \pi} = -1\]
to simplify the integrand

Answer 5

Thanks a lot. I got it.