What is $\int_0^{2 \pi} \sin^{2009}( x) ~~dx$ What is
$$\displaystyle \int_0^{2 \pi} \sin^{2009}(x) ~~dx$$
If it is $0$, how so? Then shouldn't $\sin x$ be also $0$ and not -$\cos x$ since the top and bottom cancel out?
 A: Because $\sin (\pi + x) = -\sin (\pi - x)$, we can make a substitution $u = x - \pi$.  Then we have the integral of an odd function (composition of two odd functions $\sin x$ and $x^{2009}$) between $-\pi$ and $\pi$ and so the area is $0$.
A: The integral evaluates to $0$, which can be intuited from the following argument: $\sin(x)$ is oscillatory and $2\pi$-periodic, so that there is equal area above and below the $x$-axis over the interval $[0,2\pi]$; this applies for all odd powers of $\sin(x)$. Moreover, this applies for all odd powers of $\cos(x)$. As long as you integrate over the period of these oscillating functions (their odd powers that is), the areas above and below the $x$-axis will cancel, yielding $0$.
A: Use the reduction formula to study the recurrence relation involved $$\int \sin^{2n+1}x =\cfrac{\cos x \sin^{2n} x}{2n+1}+\cfrac{2n}{2n+1}\cdot\int \sin^{2n-1}x \ dx +c $$
and $\displaystyle \int_0^{2 \pi} \sin(x) dx =\left[-\cos x \right]^{2\pi}_0= 0$  and not just $- \cos x$
