Finding Fourier Series of f(x)=√(x2) -pi/2<x<pi/2

[Evilavatar2]

Homework Statement

Find the Fourier series of the function
f(x) =√(x²) -pi/2<x<pi/2 , with period pi

Homework Equations

The Attempt at a Solution

I have tried attempting the question, but couldn't get the answer. uploaded my attempted qns with the picture attached

 

[RUber]

Your work looks solid at first glance...maybe an algebraic error, I'll go back and look at the details soon.
Remember, your final form should be something like:
$f(x) = a_0 + \sum_{n=1}^\infty a_n \cos 2n x$

 

[RUber]

You were right to use the half-period formula and notice it was an even function. You did not write it like one f(x) = |x|, not f(x) = x.
For $a_0$, it looks like you did 1/2pi, instead of 1/pi for the full period of pi. On the half-period formula, you want to double the result to get the full period, so you should end up with 2/pi as your coefficient on that integral
$a_0 = \frac2\pi \int_0^{\pi/2} f(x) \, dx .$
For your $a_n$ terms, your coefficient was incorrect as well, since it should be (L was correctly identified as pi/2).
$a_n = \frac2L \int_0^{L} f(x)\cos\left(\frac{nx}{L} \right)\, dx .$
Other than those coefficients, your integration by parts looks to be done correctly from what I can make out.
**edit** you missed a negative sign in the sine integral. Integral of sin(x) dx = - cos x. **end of edit**
Let me know if you are still having trouble.

 

[Evilavatar2]

Thanks a lot for helping. I have decided to redo the whole question and corrected the ao. But i can't find the mistake that you pointed out for the integral sin(x)dx= -cos x. Also my answer is still incorrect.

 

[RUber]

I must have lost track of the negative sign somewhere--your new solutions seem to be in the right neighborhood.
Look at the coefficient on your $a_n$ integral. It is the same as the one you used on your $a_0$ integral. It should be twice as big.