Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 40

You need to use the substitution cos t = u , such that:
cos t = u => -sin t dt = du => sin t dt = -du
Replacing the variable, yields:
int_0^pi e^(cos t)*sin (2t)dt = 2int_0^pi e^(cos t)*sin t*cos t dt
2int_0^pi e^(cos t)*sin t*cos t dt = -2int_(u_1)^(u_2) u*e^u*du
You need to use the integration by parts such that:
int fdg = fg - int gdf
f = u => df = du
dg = e^u => g = e^u
-2int_(u_1)^(u_2) u*e^u*du = -2u*e^u|_(u_1)^(u_2) + 2int_(u_1)^(u_2) e^u du
-2int_(u_1)^(u_2) u*e^u*du = (-2u*e^u + 2e^u)|_(u_1)^(u_2)
Replacing back the variable, yields:
int_0^pi e^(cos t)*sin (2t)dt = (-2cos t*e^(cos t) + 2e^(cos t))|_0^(pi)
Using the fundamental theorem of integration, yields:
int_0^pi e^(cos t)*sin (2t)dt = (-2cos pi*e^(cos pi) + 2e^(cos pi) + 2cos 0*e^(cos 0) - 2e^(cos 0))
int_0^pi e^(cos t)*sin (2t)dt = (2e^(-1) + 2e^(-1) + 2e - 2e)
int_0^pi e^(cos t)*sin (2t)dt = 4/e
Hence, evaluating the integral, using substitution, then integration by parts, yields int_0^pi e^(cos t)*sin (2t)dt = 4/e.