Question

How do you integrate `int e^xcosx` by integration by parts method?

Answer 1

By integrating by parts twice:

`\frac{e^x(cos(x) + sin(x))}{2}`

Explanation:

We will use the fact that `e^x` remains unchanged when integrated, while the cosine function has the following derivation loop:

`cos(x)-> -\sin(x) -> -\cos(x) -> sin(x) -> cos(x) -> \ldots`

In particular, we're interested in the fact that deriving the cosine function twice means to invert its sign.

So, after a first integration by parts, we have

`int e^xcos(x) = e^x cos(x) - int e^x(-sin(x)) =`
`= e^x cos(x) + int e^xsin(x)`

Now integrate `e^xsin(x)` by parts again: we have

`int e^xsin(x) = e^xsin(x) - \int e^xcos(x)`

Putting all the pieces together, we have

`int e^xcos(x) = e^x cos(x) + e^x sin(x) - \int e^xcos(x)`

As you can see, the (same) integral appears on both sides, this means that we can add it to both sides to get

`2int e^xcos(x) = e^x cos(x) + e^x sin(x)`

and thus solve for it:

`int e^xcos(x) = \frac{e^x cos(x) + e^x sin(x)}{2} = \frac{e^x(cos(x) + sin(x))}{2}`

Answer 2

`int e^x cosx dx = (e^x(cosx+sinx))/2 +C`

Explanation:

As `d(e^x) = e^xdx`, we can integrate by parts int the following way:

`int e^x cosx dx = int cosx d(e^x)`

`int e^x cosx dx = e^xcosx- int e^x d(cosx)`

`int e^x cosx dx = e^xcosx+ int e^x sinx dx`

Integrate by parts again:

`int e^x cosx dx = e^xcosx+ int sinx d(e^x)`

`int e^x cosx dx = e^xcosx+e^xsinx- int e^x d(sinx)`

`int e^x cosx dx = e^xcosx+e^xsinx - int e^x cosx dx`

The integral now appears on both sides of the equation and we can solve for it:

`2int e^x cosx dx = e^xcosx+e^xsinx +C`

`int e^x cosx dx = (e^x(cosx+sinx))/2 +C`