How do you integrate #(e^x)(cosx) dx#?

Question

Christopher Allers · Accepted Answer

This integral is cyclic and requires two iterations using the integration by parts theorem, which states:

#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#

We can assume that #f(x)=cosx# and #g'(x)dx=e^xdx#,

#f'(x)dx=-sinxdx#

#g(x)=e^x#,

so:

#I=e^xcosx-inte^x(-sinx)dx=#

#=e^xcosx+inte^xsinxdx=(1)#.

And now once more:

#f(x)=sinx# and #g'(x)dx=e^xdx#,

#f'(x)dx=cosxdx#

#g(x)=e^x#.

So:

#(1)=e^xcosx+[e^xsinx-inte^xcosxdx]=#

#=e^xcosx+e^xsinx-inte^xcosxdx#.

At last, we are able to write:

#I=e^xcosx+e^xsinx-IrArr#

#2I=e^xcosx+e^xsinx-IrArr#

#I=e^x/2(cosx+sinx)+c#.

Charles Arocha · Answer

undefined To integrate $ e^x \cdot \cos(x) \, dx $, you can use integration by parts method. Let $ u = e^x $ and $ dv = \cos(x) \, dx $.

$$ du = e^x \, dx $$
$$ v = \int \cos(x) \, dx = \sin(x) $$

Now, apply the integration by parts formula:

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

$$ \int e^x \cdot \cos(x) \, dx = e^x \cdot \sin(x) - \int \sin(x) \cdot e^x \, dx $$

This results in:

$$ \int e^x \cdot \cos(x) \, dx = e^x \cdot \sin(x) - \int e^x \cdot \sin(x) \, dx $$

Now you have another integral to solve. You can integrate it using integration by parts again or by using a combination of methods such as substitution and integration by parts.