# How do you integrate #int e^xcos(2x)# by parts?

We integrate the resulting integral by parts again:

So if we name:

we get the following equation:

so that:

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To integrate ( \int e^x \cos(2x) ) by parts, we use the integration by parts formula, which states:

[ \int u , dv = uv - \int v , du ]

Let ( u = e^x ) and ( dv = \cos(2x) , dx ). Then, ( du = e^x , dx ) and ( v = \frac{1}{2} \sin(2x) ).

Now, we apply the integration by parts formula:

[ \int e^x \cos(2x) , dx = e^x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot e^x , dx ]

The integral on the right-hand side is similar to the original integral but with ( e^x ) as ( u ) and ( \sin(2x) , dx ) as ( dv ).

So, we repeat the integration by parts:

[ u = e^x, \quad dv = \sin(2x) , dx ] [ du = e^x , dx, \quad v = -\frac{1}{2} \cos(2x) ]

Now, we plug these into the formula:

[ \int e^x \cos(2x) , dx = e^x \cdot \frac{1}{2} \sin(2x) - \left( -\frac{1}{2} e^x \cos(2x) - \int -\frac{1}{2} \cos(2x) \cdot e^x , dx \right) ]

Simplify:

[ \int e^x \cos(2x) , dx = \frac{1}{2} e^x \sin(2x) + \frac{1}{2} \int e^x \cos(2x) , dx ]

Now, we can isolate the integral term:

[ \frac{1}{2} \int e^x \cos(2x) , dx = \frac{1}{2} e^x \sin(2x) ]

Multiply both sides by ( 2 ):

[ \int e^x \cos(2x) , dx = e^x \sin(2x) + C ]

Where ( C ) is the constant of integration. This is the final result of integrating ( \int e^x \cos(2x) ) by parts.

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