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

Question

Samantha Adkison · Accepted Answer

#int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C#

As #d(e^x) = e^xdx# we can integrate by parts as:

#int e^x cos(2x) dx = int cos(2x) d(e^x) = e^xcos2x + 2int e^xsin(2x)dx#

We integrate the resulting integral by parts again:

#int e^xsin(2x)dx = int sin(2x)d(e^x) = e^xsin2x -2int e^xcos(2x)dx#

So if we name:

#I = int e^x cos(2x) dx #

we get the following equation:

#I = e^xcos2x+2e^xsin2x-4I#

and solving for #I#:

#I= 1/5(e^xcos2x+2e^xsin2x)#

so that:

#int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C#

Paisley Johnson · Answer

undefined 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.