# Using the method of integration for parts show that ∫e^x sin(x)dx = 1/2sin(x)e^x - 1/2cos(x)e^x + constant

You should remember the formula of integration by parts such that:

int udv = uv  - int vdu

Considering u = sin x  and dv = e^x dx  yields:

u = sin x => du = cos x dx

dv = e^x dx => v = e^x

int e^x sin x dx = e^x sin x - int cos x e^x dx

Using integration by parts again to evaluate int cos x e^x dx   yields:

u = cos x => du = -sin x dx

dv = e^x dx => v = e^x

int cos x e^x dx = e^x cos x + int e^x sin x dx

You should come up with the following notation for int e^x sin x dx  such that:

I = int e^x sin x dx

I = e^x sin x - (e^x cos x + I) => I = e^x sin x - e^x cos x- I

Moving the terms that contain I  to the left side yields:

I + I = e^x sin x - e^x cos x => 2I = e^x sin x - e^x cos x

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

Substituting back int e^x sin x  dx forI  yields:

int e^x sin x dx = (1/2)e^x sin x - (1/2)e^x cos x + c

Hence, evaluating the given integral using parts yields int e^x sin x dx = (1/2)e^x sin x - (1/2)e^x cos x + c .

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