Integrate e^xsinx dx

Question

I did integration by parts and then got stuck I need an understanding of where the 2 comes from?

Answer 1

IBP #1

integral [ e^x sin x ]

u = e^x dv = sin x

du = e^x v = -cos x

- e^x * cos x + integral [ -cos x * e^x ]

IBP #2

U = e^x dv = -cos x

du = e^x V = sin x

- e^x * cos x + sin x * e^x - integral [ sin x * e^x]

So the ORIGINAL integral is equal to itself PLUS some other functions:

In summary

integral [ e^x sin x ] = - e^x * cos x + sin x * e^x - integral [ sin x * e^x]

moving the integral on the right side to the left side,

2 * integral [ e^x sin x] = - e^x * cos x + sin x * e^x

integral [ e^x sin x] = [sin x e^x - e^x cos x]/2 = (1/2) e^x (sin x - cosx)

Check by differentiation:

(1/2) [ e^x( cos x + sinx) + e^x ( sin x - cos x) ]

(1/2) e^x ( cos x + sin x + sin x - cos x )

(1/2) e^x (2 sin x)

e^x sin x

THerefore, the anti-derivative in bold above is correct