# 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?

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The integrand is $2\pi$-periodic so you can integrate over any interval of length $2\pi$ and get the same answer. Integrate over $[-\pi, \pi]$. Your integrand is an odd function so you get $0$. –  Stefan Smith Apr 18 '13 at 0:02

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 $0$.

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What? I don't understand. Why did you add pi to the x inside of sin? You can't plug in 2pi and 0 until you integrate sin^2009 (x) dx.... –  Jo Mo Apr 17 '13 at 4:01
If you know integration by substitution, try applying the substitution rule $u = x - \pi$ for your integral and see what happens. –  muzzlator Apr 17 '13 at 4:04
I am wondering how you got x-pi in there in the first place –  Jo Mo Apr 17 '13 at 5:05
There is an antisymmetry of $\sin x$ about $x=\pi$ and so $u = x - \pi$ allows us to move the antisymmetry to $u=0$ –  muzzlator Apr 17 '13 at 5:34

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 $- \cos x$

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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 $0$.

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I wouldn't strictly say because it's $2 \pi$ periodic. More that it is odd about $x = \pi$. –  muzzlator Apr 17 '13 at 3:52
@muzzlator : I'd say though that this is a nice intuitive look at what's going, useful for the OP. –  Coffee_Table Apr 17 '13 at 4:04
I think the intuitive feel which you have and you wanted to share with the OP wasn't properly articulated. You said that there is equal area above and below the axis for $\sin x$ because $\sin x$ is $2 \pi$-periodic. But there are many $2 \pi$-periodic functions which don't have equal area below and above the axis. So while what you said is true, the justification which you wrote for why it was true wasn't, I think, what you yourself pictured. –  muzzlator Apr 17 '13 at 4:22
@muzzlator : Point taken. I've modified my answer to reflect the informal nature of my argument. –  Coffee_Table Apr 17 '13 at 20:21