[Math] integral of sin(x) to the power 2014

Question

For a course in Complex Analysis we're tasked to find the integral of
\begin{align*}
\int_0^{2 \pi} (\sin\theta)^{2014} d \theta
\end{align*}
but I'm a bit stumped so far on how to do this.

What I've done so far:

• First I tried to replace $ \sin \theta$ by $\frac{e^{i \theta} – e^{-i \theta}}{2i}$, but you still have this weird term $(e^{i \theta} – e^{-i \theta})^{2014}$ to deal with, so I guess that's not the correct way.
• Secondly I thought about intepreting this as the imaginary part of $e^{i \theta}$, but I get stuck on the 2014 power, so this method presents problems as well.

If anyone could point me in the right direction I would be very grateful, thank you!

Answer 1

HINT :

Let $$ I_n=\int\sin^n ax\ dx, $$ then using integrating by parts, it is not difficult to obtain $$ anI_n=-\sin^{n-1} ax\cos ax+a(n-1)I_{n-2}. $$ It is called integration by reduction formula.

Note that, if we use integration by reduction formula for $3$ or $4$ steps, you will see the pattern that leads to the result $$ \int_0^{2\pi}\sin^{2014} x\ dx=2\pi\cdot\frac{2013!!}{2014!!}. $$ I just don't get it why my answer got vote down without double-checking first. Maybe that was just an unintentionally mistake. just try to stay positive. :)

You can also write: $$ \sin ax=\frac{e^{iax}-e^{-iax}}{2i}\quad\Rightarrow\quad\sin^n ax=\left(\frac{e^{iax}-e^{-iax}}{2i}\right)^n $$ and then expand the term $\left(e^{iax}-e^{-iax}\right)^n$ using Binomial theorem. Perhaps using Binomial theorem would be a bit longer than using integration by reduction formula.

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