# Calc1.57

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$\int e^{x}\sin x\,dx\,$

To do this integral, we will use the method of integration by parts which tells us

$\int u\,dv=uv-\int v\,du\,$.

For this specific integral, the technique is to create an equation by using integration by parts twice and then solve for the value of the integral. The exact same integral will pop up after we do our second integration by parts at which time we can move it to the other side of the equation.

$u=\sin x\,$

$du=\cos x\,dx\,$

$dv=e^{x}\,dx\,$

$v=e^{x}\,$

So, our integral becomes

$\int e^{x}\sin x\,dx=e^{x}\sin x-\int e^{x}\cos x\,dx\,$

Do integration by parts again with this new integral. Technically, $u\,$, $du\,$, $v\,$, and $dv\,$ all are already assigned values but it is common practice to use these same variables again for this new integral. However, keep in mind that they have nothing to do with the previous $u\,$, $du\,$, $v\,$, and $dv\,$.

$u=\cos x\,$

$du=-\sin x\,dx\,$

$dv=e^{x}\,dx\,$

$v=e^{x}\,$

This gives us

$\int e^{x}\sin x\,dx=e^{x}\sin x-\left[e^{x}\cos x+\int e^{x}\sin x\,dx\right]\,$

Simplifying the equation gives us

$\int e^{x}\sin x\,dx=e^{x}\sin x-e^{x}\cos x-\int e^{x}\sin x\,dx\,$

Now, we can add $\int e^{x}\sin x\,dx\,$ to both sides of the equation and divide by $2\,$ to get the value of the integral.

$\int e^{x}\sin x\,dx={\frac {e^{x}}{2}}(\sin x-\cos x)+C\,$