Question

What is the antiderivative of `(e^x)sinx`?

Answer 1

`inte^xsin(x)dx=(e^x(sin(x)-cos(x)))/2+C`

Explanation:

Use integration by parts:

`intudv=uv-intvdu`

So, for

`intcolor(red)(e^x)color(blue)(sin(x))color(red)dx`

We should let:

`color(blue)(u=sin(x))" "=>" "du=cos(x)dx`

`color(red)(dv=e^xdx)" "=>" "intdv=inte^xdx" "=>" "v=e^x`

Plugging these into the integration by parts formula, we see that

`inte^xsin(x)dx=e^xsin(x)-intcolor(green)(e^x)color(purple)(cos(x))color(green)dx`

Use integration by parts for the second time:

`color(purple)(u=cos(x))" "=>" "du=-sin(x)dx`

`color(green)(dv=e^xdx)" "=>" "intdv=inte^xdx" "=>" "v=e^x`

Thus, we obtain

`inte^xsin(x)dx=e^xsin(x)-[e^xcos(x)-inte^x(-sin(x))dx]`

`inte^xsin(x)dx=e^x(sin(x)-cos(x))-inte^xsin(x)dx`

Add the integral to both sides and solve for it:

`2inte^xsin(x)dx=e^x(sin(x)-cos(x))`

`inte^xsin(x)dx=(e^x(sin(x)-cos(x)))/2+C`