What is the antiderivative of (e^x)sinx(ex)sinx?
1 Answer
Explanation:
Use integration by parts:
intudv=uv-intvdu∫udv=uv−∫vdu
So, for
intcolor(red)(e^x)color(blue)(sin(x))color(red)dx∫exsin(x)dx
We should let:
color(blue)(u=sin(x))" "=>" "du=cos(x)dxu=sin(x) ⇒ du=cos(x)dx
color(red)(dv=e^xdx)" "=>" "intdv=inte^xdx" "=>" "v=e^xdv=exdx ⇒ ∫dv=∫exdx ⇒ v=ex
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∫exsin(x)dx=exsin(x)−∫excos(x)dx
Use integration by parts for the second time:
color(purple)(u=cos(x))" "=>" "du=-sin(x)dxu=cos(x) ⇒ du=−sin(x)dx
color(green)(dv=e^xdx)" "=>" "intdv=inte^xdx" "=>" "v=e^xdv=exdx ⇒ ∫dv=∫exdx ⇒ v=ex
Thus, we obtain
inte^xsin(x)dx=e^xsin(x)-[e^xcos(x)-inte^x(-sin(x))dx]∫exsin(x)dx=exsin(x)−[excos(x)−∫ex(−sin(x))dx]
inte^xsin(x)dx=e^x(sin(x)-cos(x))-inte^xsin(x)dx∫exsin(x)dx=ex(sin(x)−cos(x))−∫exsin(x)dx
Add the integral to both sides and solve for it:
2inte^xsin(x)dx=e^x(sin(x)-cos(x))2∫exsin(x)dx=ex(sin(x)−cos(x))
inte^xsin(x)dx=(e^x(sin(x)-cos(x)))/2+C∫exsin(x)dx=ex(sin(x)−cos(x))2+C
