proving that $|\int_0^{2\pi}f(t)\sin(nt)dt| \le \frac{4}{n^2}$ I've been trying for hours to prove that $|\int_0^{2\pi}f(t)\sin(nt)dt| \le \frac{4}{n^2}$ according the following question, with no luck.
The question is: 
Given that $f(x)$ has a continuous second derivative in $[0,2\pi]$, $f(0)=f(2\pi)$ and $|f''(x)| \le 1$ for every $x \in [0,2\pi]$. Prove that $|\int_0^{2\pi}f(t)\sin(nt)dt| \le \frac{4}{n^2}$ for every natural n.
What I tried to do: I tried to "get to" the second derivative using integration by parts. Since we know that $f(0)=f(2\pi)$, I tried to take an integral from 0 to 1, and from 1 to $2\pi$ and since we know that $|f''(x)| \le 1$, we know that the absolute value will always be a fracture. But I couldn't get to prove that it will always be less than $\frac{4}{n^2}$. 
Hope you could help me with it.
Another note, while I researched I saw that similar problems are solved using fourier transformations - this is not a fourier question.
 A: Well, we know that:
$$\int\text{f}\left(t\right)\cdot\text{g}\space'\left(t\right)\space\text{d}t=\text{f}\left(t\right)\cdot\text{g}\left(t\right)-\int\text{f}\space'\left(t\right)\cdot\text{g}\left(t\right)\space\text{d}t\tag1$$
So, when $\text{f}\left(t\right)=\text{f}\left(t\right)$ and $\text{g}\space'\left(t\right)=\sin\left(\text{n}\cdot t\right)$, we get:
$$\int_0^{2\pi}\text{f}\left(t\right)\cdot\sin\left(\text{n}\cdot t\right)\space\text{d}t=\frac{1}{\text{n}}\cdot\left\{\text{f}\left(0\right)-\cos\left(2\pi\cdot\text{n}
\right)\cdot\text{f}\left(2\pi\right)+\int_0^{2\pi}\text{f}\space'\left(t\right)\cdot\cos\left(\text{n}\cdot t\right)\space\text{d}t\right\}\tag2$$
Now, we know that $\text{f}\left(0\right)=\text{f}\left(2\pi\right)$ and $\text{n}\in\mathbb{N}^+$, so we get:
$$\int_0^{2\pi}\text{f}\left(t\right)\cdot\sin\left(\text{n}\cdot t\right)\space\text{d}t=\frac{1}{\text{n}}\cdot\left\{\int_0^{2\pi}\text{f}\space'\left(t\right)\cdot\cos\left(\text{n}\cdot t\right)\space\text{d}t\right\}\tag3$$
Using IBP again:
$$\frac{1}{\text{n}}\cdot\left\{\int_0^{2\pi}\text{f}\space'\left(t\right)\cdot\cos\left(\text{n}\cdot t\right)\space\text{d}t\right\}=$$
$$\frac{1}{\text{n}}\cdot\left\{\left[\text{f}\space'\left(t\right)\cdot\frac{\sin\left(\text{n}\cdot t\right)}{\text{n}}\right]_0^{2\pi}-\int_0^{2\pi}\text{f}\space''\left(t\right)\cdot\frac{\sin\left(\text{n}\cdot t\right)}{\text{n}}\space\text{d}t\right\}=$$
$$-\frac{1}{\text{n}^2}\int_0^{2\pi}\text{f}\space''\left(t\right)\cdot\sin\left(\text{n}\cdot t\right)\space\text{d}t\tag4$$
