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
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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 for`I` 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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