<s>[INST] Sine of argument with large n approximation I have worked an integral and reduced the integral to
$$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$
I want to show that for $$n\rightarrow \infty$$ the above equation reduces to
$$\frac{1}{2}$$
Evidently, this means the $2$ sine functions must cancel each other. But what is a good way to do this? Large $n$ results in sine toggling between $-1$ and $1$. Notice that the either sine function will have a sign opposite to the other.
 [/INST] First notice that
$$\begin{align}
\sin(\frac{3n\pi}{2})&=\sin(\frac{n\pi}{2}+n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+\cos(\frac{n\pi}{2})\sin(n\pi) \\
&=\sin(\frac{n\pi}{2})\cos(n\pi)+0 \\
&=(-1)^n\sin(\frac{n\pi}{2})
\end{align}$$
and hence
$$\sin(\frac{n\pi}{2})-\sin(\frac{3n\pi}{2})=(1-(-1)^n)\sin(\frac{n\pi}{2})$$
The above expression is $0$ or $2$ depending on $n$ is even or odd. Now, you can find a bound easily
$$0 \le (1-(-1)^n)\sin(\frac{n\pi}{2}) \le 2$$
And finally 
$$\frac{1}{2} \le \frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}\le \ \frac{1}{2} + \frac{1}{ n \pi}$$
Now, you can guess what happens next! :)
</s>