`int_0^pi e^(cos(t)) sin(2t) dt` First make a substitution and then use integration by parts to evaluate the integral

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) +...

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.`