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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