# Evaluate the `int e^x*sin xdx` Please explain as you solve thanks

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`int e^xsinx dx`

To evaluate, use integration by parts. The formula is `int udv = uv - intvdu` .

So let,

`u=e^x` and `dv=sinx dx`

`du=e^xdx` `v=int sinx dx = -cos x`

Substituting this to the formula yields,

`int e^x sinx dx = e^x(-cosx) - int - cosx e^x dx `

`= -e^xcosx + int e^x cosxdx`

To evaluate the integral part, use integration by parts again. So let,

`u=e^x` and `dv=cosxdx`

`du=e^xdx` `v= int cosxdx = sinx`

We then have,

`inte^xsinxdx=-e^xcosx+[e^xsinx-int sin xe^xdx]`

`inte^xsinx dx=-e^xcosx+e^xsinx-int e^xsinx dx`

Take note that both sides have the same integral. To simplify, add both sides by` inte^xsinx dx` .

`2inte^xsinx dx = -e^xcosx+e^xsinx`

To solve for `int e^xsinx dx` , divide both sides by 2.

`int e^xsinxdx = (-e^xcosx + e^xsinx)/2`

`inte^xsinxdx = (e^x (sinx - cosx))/2`

--------------------------------------------------------------------------------------- **Since we have an indefinite integral , therefore:`inte^xsinx dx = (e^x(sinx-cosx))/2 + C` .**