Using Fourier series to determine if a function is odd or even Determine if the following function is odd or even using Fourier series. 
$$f(x)=x^5\sin x$$
If a function is even then $b_n=0$ and you have to evaluate $a_n=\frac 2 \pi \int_0^\pi f(x) \cos nx \, dx$  
And if a function is odd then $a_n=0$ and you have to evaluate $b_n=\frac 2 \pi \int_0^\pi f(x)\sin nx \, dx$
However, this means that I have to do integration by parts five times. Is there a more efficient method to determine if the function is even or odd? 
 A: You do not need complicated tools to determine whether a function $f(x)$ is odd, even or neither-nor.
A function $f(x)$ is even, if and only if the equation $f(-x)=f(x)$ holds for every $x$ for which both $f(x)$ and $f(-x)$ are defined and it is odd if and only if the equation $f(-x)=-f(x)$ holds for every $x$ for which both $f(x)$ and $f(-x)$ are defined. 
In your example, we have $f(-x)=(-x)^5\sin(-x)=(-x^5)\cdot (-sin(x))=x^5\cdot sin(x)=f(x)$, so $f$ is even.
A: If a function is even then $b_n=0$ and you have to evaluate $b_n=\frac {2}{\pi}∫_{-\pi}^{\pi} f(x)\sin nx\ dx$
Note, you do not have to find any of the values of $a_n.$  You need to show that all the values of $b_n$ equal 0.
And, note the limits of integration.
Similarly: If a function is odd then $a_n=0$ and you have to evaluate $a_n=\frac {2}{\pi}∫_{-\pi}^{\pi} f(x)\cos nx\ dx$
$b_n=\frac {2}{\pi}∫_{-\pi}^{\pi} x^5\sin nx\ dx$
Now, rather than evaluating, and performing integration by parts 5 times.
I suggest you show that: 
$a_n=\frac {2}{\pi}∫_{-\pi}^{\pi} \cos nx\ dx = 0$
and
b_n=\frac {2}{\pi}∫_{-\pi}^{\pi} x\sin nx\ dx = 0$
Then use induction to show that
$a_n=\frac {2}{\pi}∫_{-\pi}^{\pi} x^m\cos nx\ dx = 0$ when $m$ is even 
and 
$b_n=\frac {2}{\pi}∫_{-\pi}^{\pi} x^m\sin nx\ dx = 0$ when $m$ is odd
