What is the antiderivative and how do you integrate sin(x)e^x dx?

Answer 1

e^x (sin(x) - cos(x)) / 2

Try integration by parts twice.

u = sin(x) ; dv = e^x
du = cos(x) ; v = e^x

∫sin(x)e^x dx
= e^x sin(x) - ∫cos(x)e^x dx

Now we need to figure this second integral:
∫cos(x)e^x dx
u = cos(x) ; dv = e^x
du = -sin(x) ; v = e^x
∫cos(x)e^x dx = e^x cos(x) - ∫-sin(x)e^x dx
= e^x cos(x) + ∫sin(x)e^x dx

Substituting this into the first, we get:

∫sin(x)e^x dx
= e^x sin(x) - [e^x cos(x) + ∫sin(x)e^x dx]
= e^x sin(x) - e^x cos(x) - ∫sin(x)e^x dx
or
2 ∫sin(x)e^x dx = e^x sin(x) - e^x cos(x)
∫sin(x)e^x dx = e^x (sin(x) - cos(x)) / 2

Answer 2

By parts
Int e^x*sin(x)dx= e^x sin(x) -Int e^x cos(x) and

Int e^x cos(x) = e^x cos +Int e^x sinx so

Int e^x sin x = e^x sinx - e^x cos x - Int e^x sin x.

So Int e^x sin x = 1/2e^x(sinx-cosx)