Integrate e^x sinx dx?

Integration by parts : ∫u dv = u v - ∫v du

∫e^x sinx dx
= ∫e^x d(-cosx)
= e^x (-cosx) - ∫(-cosx) d(e^x)
= -e^x cosx + ∫cosx e^x dx
= -e^x cosx + ∫e^x (cosx dx)
= -e^x cosx + ∫e^x (d sinx)
= -e^x cosx + [e^x sinx - ∫sinx d(e^x)]
= -e^x cosx + e^x sinx - ∫e^x sinx dx
= e^x sinx - e^x cosx - ∫e^x sinx dx
= e^x (sinx - cosx) - ∫e^x sinx dx

∫e^x sinx dx = e^x (sinx - cosx) - ∫e^x sinx dx
2 ∫e^x sinx dx = e^x (sinx - cosx)

∫e^x sinx dx = e^x (sinx - cosx) / 2

Integrate by parts twice:
u1 = sin(x) dx, v1 = e^x dx,
du1 = cos(x) dx, v1 = e^x.
Integral of u dv = uv - int v du
= e^x sin(x) - int e^x cos(x) dx.
Now let u2 = -cos(x), dv2 = e^x dx,
du2 = sin(x) dx, v2 = e^x.

Now you have:
Integral of e^x sin(x) dx is
e^x sin(x) - e^x cos(x) - integral of e^x sin(x) dx.

So add the integral of e^x sin(x) to both sides, obtaining:
2*integral e^x sin(x) dx = e^x (sin(x) - cos(x)), and so:
integral e^x sin(x) = (1/2)*e^x [sin(x) - cos(x)].

∫e^x sinx dx. Integrate by parts.

Let u = e^x, dv = sinx dx. Then du = e^x dx, v = -cosx

-e^x cosx + ∫e^x cosx dx. Integrate by parts again.

Let s = e^x, dt = cosx dx. Then ds = e^x dx, t = sinx

-e^x cosx + e^x sinx - ∫e^x sinx dx = ∫e^x sinx dx

This appears to be circular reasoning, but we can solve for the integral algebraically.

∫e^x sinx dx = ½e^x(sinx - cosx) + C

Here's the solution on youtube:

https://search.yahoo.com/yhs/search?p=e%5Ex+sinx+dx&ei=UTF-8&hspart=mozilla&hsimp=yhs-002