Question

Evaluate ∫ e−stsint dt?

Answer 1

start by using integration by parts two times, and then solve algebraically for the integral you want:

∫e−stsin(2t)dt=−12e−stcos(2t)−s2∫e−stcos(2t)dt=−12e−stcos(2t)−s2(12e−stsin(2t)+s2∫e−stsin(2t)dt)=−12e−stcos(2t)−s4e−stsin(2t)−s24∫e−stsin(2t)dt

So, that's two applications of integration by parts. Now you can add that last term back over to the left-hand side:

∫e−stsin(2t)dt+s24∫e−stsin(2t)dt=−12e−stcos(2t)−s4e−stsin(2t)

Multiplying everything by 4 and factoring appropriately, we obtain:

(4+s2)∫e−stsin(2t)dt=−e−st(2cos(2t)+ssin(2t))

or,

∫e−stsin(2t)dt=−e−st(2cos(2t)+ssin(2t))4+s2