Integration by Parts: Solve ∫cos(x)cos(kx)dx

[8700]

Homework Statement

Hello!

I am having some trouble solving this integral by parts. I hope someone can help me.

$\int \cos(x)cos(kx) dx$

It is need for a Fourier series

Homework Equations

I am using this definition:

$\int f(x)g(x) dx = f(x)G(x)-\int f'(x)G(x) dx$

since its an even function i need to find $a_k$ and $a_0$, I've found $a_0$ no worries about that.

The Attempt at a Solution

[/B]

$\int f(x)g(x) dx = \frac{cos(x)sin(kx)}{k} - \int \frac{-sin(x)sin(kx)}{k}$

I can't remember/see the next rule of integration when having two sin functions. Or am i doing something wrong?

 

[BvU]

Any particular knowledge about k you can share with us ?

Oh, and: use the template !

 

[8700]

k is just an integer that's about it.

I'll remember that next time.

 

[BvU]

So what do you do when k = 1 ?
Next question: what do you do when k = 2 ?
And you can guess the third question, right ?

 

[8700]

I need it for a Fourier series where the inteval is from [0, $\pi$ ]. f(x)=cos(x) so it is an even function which means i need to calculate $a_0$ and $a_k$.

It is $a_k$ who i am trying to calculate right now by parts, but i am stuck at it.

 

[BvU]

Ah, so it concerns integrals with definite limits. Also something that would have been clearly described under 1.
And you would have provided some relevant equations under 2. e.g. $\sin k\pi=0$, etc.

Never too late to catch up!

Homework Statement

Homework Equations

The Attempt at a Solution

Have to go now, perhaps others can take over...

 

[8700]

yeah sorry!

Actually that is what confuses me. if $\sink\pi=0$ then there is only $\frac{cos(x)sin(kx)}{k}$ left? Does that mean that this is my $a_k$??

 

[8700]

so when taking the interval into account i get the result $a_k=\frac{-sin(k\pi)}{k}$ which is wroooong

 

[gneill]

The Attempt at a Solution

[/B]

$\int f(x)g(x) dx = \frac{cos(x)sin(kx)}{k} - \int \frac{-sin(x)sin(kx)}{k}$

I can't remember/see the next rule of integration when having two sin functions. Or am i doing something wrong?

The new integral looks very much like the initial one, only using sin() rather than cos(). Your instinct should be to apply Integration by Parts again, leading to something of the form:

f(x) = g(x) - c*f(x)

from which you can isolate f(x).