integration by complex exponentials

I have seen that one can use the euler's formula to complexify the integration, and integrate functions like $\displaystyle e^{nx} \cos mx dx$ easily.(The method has a wikipedia page:Integration using Euler's formula - Wikipedia, the free encyclopedia)

How can i do $\displaystyle \int \frac{dx}{13+3\cos x+4\sin x}$ using complex exponentials.

PS:I know this can be easily done with standard trig substitutions and might turn out to be longer using complex exponentials, but I want to know if and how can it be done with this method ...(Nod)

Re: integration by complex exponentials

Quote:

Originally Posted by

**earthboy** I have seen that one can use the euler's formula to complexify the integration, and integrate functions like $\displaystyle e^{nx} \cos mx dx$ easily.(The method has a wikipedia page:

Integration using Euler's formula - Wikipedia, the free encyclopedia)

How can i do $\displaystyle \int \frac{dx}{13+3\cos x+4\sin x}$ using complex exponentials.

PS:I know this can be easily done with standard trig substitutions and might turn out to be longer using complex exponentials, but I want to know if and how can it be done with this method ...(Nod)

$\displaystyle cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$ and $\displaystyle sin(x)= \frac{e^{ix}- e^{-ix}}{2}$ so that integral, in complex exponentials, is $\displaystyle \int\frac{dx}{13+ \frac{3e^{ix}+ 3e^{-ix}}{2}+ \frac{4e^{ix}- 4e^{-ix}}{2}}$

$\displaystyle = 2\int \frac{dx}{26+ 7e^{ix}- e^{ix}}$

Multiply both numerator and denominator by $\displaystyle e^{ix}$ to get

$\displaystyle 2\int \frac{e^{ix}dx}{26e^{ix}+ 7e^{2ix}- 1}$

Now let $\displaystyle u= e^{ix}$ so that $\displaystyle du= ie^{ix}dx$, $\displaystyle -i du= e^{ix}dx$ and that integral becomes

$\displaystyle -2i\int \frac{du}{7u^2+ 26u- 1}$.

That denominator can be factored (though not easily- use the quadratic formula to find the roots). Then it will be of the form

$\displaystyle -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du$

where a and b are the roots. That can be integrated by "partial fracations".

Re: integration by complex exponentials

Quote:

Originally Posted by

**HallsofIvy** $\displaystyle cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$ and $\displaystyle sin(x)= \frac{e^{ix}- e^{-ix}}{2}$

small doubt:

is $\displaystyle \sin x= \frac{e^{ix}- e^{-ix}}{2}$ or $\displaystyle \sin x= \frac{e^{ix}- e^{-ix}}{2i}$

both are seem to be used in books, but what is the difference?

Re: integration by complex exponentials

Quote:

Originally Posted by

**earthboy** small doubt:

is $\displaystyle \sin x= \frac{e^{ix}- e^{-ix}}{2}$ or $\displaystyle \sin x= \frac{e^{ix}- e^{-ix}}{2i}$

both are seem to be used in books, but what is the difference?

It's $\displaystyle \displaystyle \begin{align*} \sin{(x)} = \frac{e^{i\,x} - e^{-i\,x}}{2i} \end{align*}$.

Re: integration by complex exponentials

Quote:

Originally Posted by

**HallsofIvy**

That denominator can be factored (though not easily- use the quadratic formula to find the roots). Then it will be of the form

$\displaystyle -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du$

where a and b are the roots. That can be integrated by "partial fracations".

$\displaystyle -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du = -\frac{2i}{7} \frac{\log(u-a)-\log(u-b)}{a-b}$

where $\displaystyle a=-13-4\sqrt{11} $ and $\displaystyle b= 4\sqrt{11}-13$

how do I separate the real and imaginary part here to get the answer,especially what to do with the $\displaystyle \frac{2i}{7}$ ?

Re: integration by complex exponentials

Quote:

Originally Posted by

**earthboy** $\displaystyle -\frac{2i}{7}\int \frac{du}{(u- a)(u- b)}du = -\frac{2i}{7} \frac{\log(u-a)-\log(u-b)}{a-b}$

where $\displaystyle a=-13-4\sqrt{11} $ and $\displaystyle b= 4\sqrt{11}-13$

how do I separate the real and imaginary part here to get the answer,especially what to do with the $\displaystyle \frac{2i}{7}$ ?

Any hint on how to separate the real and the imaginary part and find the answer from here?