# Stuck in Integrating e^(i*x)cos(x)

## Main Question or Discussion Point

I'am trying to prove

$$\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}$$

Wolfram tells so http://integrals.wolfram.com/index.jsp?expr=e^(i*x)cos(x)&random=false

But I am stuck in obtaining the first term:

My step typically involved integration by parts:

let $u=e^{ix}cos(x)$ and $dv=dx$

so:

$$du=-e^{ix}sin(x)dx+icos(x)e^{ix}$$$$du=ie^{ix}(sin(x)+cos(x))dx$$$$du=ie^{2ix}dx$$

the other one is just: $v=x$

then:
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx$$

We will again do another integration by parts for the second term, so we let $u=x$ and $dv=e^{2ix}dx$ then solving further, we obtain:
$$\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}$$
plugging it back to the original problem, then doing simple distribution, we will obtain:
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )$$
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}$$
notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:
$$\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}$$

What have I gone wrong?

SteamKing
Staff Emeritus
Homework Helper
I'am trying to prove

$$\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}$$

Wolfram tells so http://integrals.wolfram.com/index.jsp?expr=e^(i*x)cos(x)&random=false

But I am stuck in obtaining the first term:

My step typically involved integration by parts:

let $u=e^{ix}cos(x)$ and $dv=dx$

so:

$$du=-e^{ix}sin(x)dx+icos(x)e^{ix}$$$$du=ie^{ix}(sin(x)+cos(x))dx$$$$du=ie^{2ix}dx$$

the other one is just: $v=x$

then:
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx$$

We will again do another integration by parts for the second term, so we let $u=x$ and $dv=e^{2ix}dx$ then solving further, we obtain:
$$\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}$$
plugging it back to the original problem, then doing simple distribution, we will obtain:
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )$$
$$\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}$$
notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:
$$\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}$$

What have I gone wrong?
You can side step a lot of the work above if you remember:

$$cos\, x = \frac{e^{ix}+e^{-ix}}{2}$$

flux!
I'm sorry, I couldn't follow your answer. But the integral is quite easy. Try using Euler's formula for e^x. Once you do ( and use a simple trigonometric identity) it's an easy
integration.
.
.
Edit: Or you can use the way in the post above mine, your convenience.

Last edited:
flux!
Mark44
Mentor
I'am trying to prove

$$\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}$$
All you really need to do is to show that ##\frac d{dx}[\frac x 2 - \frac 1 4 ie^{2ix}] = e^{ix}\cos(x)##. This isn't that hard to do.

flux!
mathman
Use streamking's approach $cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. It is now a trivial problem.