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Prove that $\int_0^{2\pi}\sin nx\,dx=\int_0^{2\pi}\cos nx\,dx=0$ for all integers $n \neq 0$.

I think I'm encouraged to prove this by induction (but a simpler method would probably work, too). Here's what I've attempted: $$\text{1.}\int_0^{2\pi}\sin x\,dx=\int_0^{2\pi}\cos x\,dx=0.\;\checkmark\\\text{2. Assume}\int_0^{2\pi}\sin nx\,dx=\int_0^{2\pi}\cos nx\,dx=0.\;\checkmark\\\text{3. Prove}\int_0^{2\pi}\sin (nx+x)\,dx=\int_0^{2\pi}\cos (nx+x)\,dx=0.\\\text{[From here, I'm lost. I've tried applying a trig identity, but I'm not sure how to proceed.]}\\\text{For the}\sin\text{integral},\int_0^{2\pi}\sin (nx+x)\,dx=\int_0^{2\pi}\sin nx\cos x\,dx+\int_0^{2\pi}\cos nx\sin x\,dx.$$

I hope I'm on the right track. In the last step, I have $\sin nx$ and $\cos nx$ in the integrals, but I'm not sure if that helps me. I would appreciate any help with this. Thanks :)


As I indicated above, it'd be great to find a way to complete this induction proof—probably by, as Arkamis said, "working it like the transcontinental railroad" with trig identities (if that's possible). I think my instructor discouraged a simple $u$-substitution, because we've recently been focused on manipulating trig identities.

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    $\begingroup$ Work it like the transcontinental railroad. Start applying trig identities to the $\sin$ integral, and others to the $\cos$ integral in parallel. Eventually, if you apply identities the right way, you should get something that matches up! $\endgroup$
    – Emily
    Oct 22, 2013 at 21:07
  • $\begingroup$ What integration techniques have you known so far? $\endgroup$ Oct 22, 2013 at 21:42
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    $\begingroup$ @Jackson : this is so trivial to do using substitution, you should find out exactly what tools you are permitted to use. Is induction required? $\endgroup$ Oct 22, 2013 at 22:13
  • $\begingroup$ @StefanSmith As far as I know, induction is not required. I'm able to use integrals of polynomials, integrals of $\sin$ and $\cos$, and trig identities. $\endgroup$
    – Jackson
    Oct 22, 2013 at 22:20
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    $\begingroup$ @StefanSmith Sorry if I caused a bit of frustration. I updated the question. $\endgroup$
    – Jackson
    Oct 25, 2013 at 2:51

7 Answers 7

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Why not just compute it directly, using a substitution? If $u = nx$, then $\frac{du}{n} = dx$ and

\begin{align*} \\\int_0^{2\pi} \sin{nx} dx &= \int_{x = 0}^{x = 2\pi} \sin{u} \frac{du}{n} \\ &= -\frac{1}{n} \cos{u} \Big|_{x = 0}^{x = 2\pi} \\ &= -\frac{1}{n} \cos nx \Big|_0^{2\pi} \\ &= \frac{1}{n} \left(\cos{2\pi n} - \cos{0}\right) \\ &= \frac{1}{n} (1 - 1) = 0 \end{align*}

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  • $\begingroup$ I'm not allowed to use $u$-substitution in my class, unfortunately—we're taking a different approach, and I wouldn't get credit for that. $\endgroup$
    – Jackson
    Oct 22, 2013 at 21:01
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I do not know if you can use $e^{ix}=\cos x+i\sin x$ but here is one solution: $$ \int_0^{2\pi}e^{inx}\,dx=\frac 1{in}e^{inx}\left|_0^{2\pi}\right.=0 $$

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  • $\begingroup$ I appreciate the answer, but this is definitely out of the scope of what I've learned and am able to apply in the class. $\endgroup$
    – Jackson
    Oct 22, 2013 at 21:11
  • $\begingroup$ @Jackson: By the way, what course are you taking now? $\endgroup$ Oct 22, 2013 at 21:41
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Hint: $\sin(x+\pi)=-\sin(x)$ and $\cos(x+\pi)=-\cos(x)$.

What does the hint mean for $\int_0^{\pi/n}\sin(nx)\,\mathrm{d}x$ and $\int_{\pi/n}^{2\pi/n}\sin(nx)\,\mathrm{d}x$ ?

For $\int_0^{\pi/n}\cos(nx)\,\mathrm{d}x$ and $\int_{\pi/n}^{2\pi/n}\cos(nx)\,\mathrm{d}x$ ?

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The substitution $t=2\pi - x$, $dt=-dx$ gives

$$\int_0^{2\pi} \sin nx\: dx = \int_{2\pi}^0 \sin (-nt) (-dt) = \int_{2\pi}^0 \sin nt\: dt = -\int_0^{2\pi} \sin nx\: dx$$

so $\int_0^{2\pi} \sin nx dx=0$. No need to evaluate the integral!

Can you find a similar trick for $\cos$?

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  • $\begingroup$ @StefanSmith Ah. It wasn't written in the question... $\endgroup$ Oct 22, 2013 at 23:25
  • $\begingroup$ @Marie Could you explain the trick for $\cos$? $\endgroup$
    – Jackson
    Oct 22, 2013 at 23:36
  • $\begingroup$ @Marie : OP edited the question, saying the instructor "discouraged a simple $u$ substitution" (see updated question). $\endgroup$ Oct 25, 2013 at 18:03
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Use the fact that $\sin$ is $2\pi$-periodic gives you

$$\int_0^{2\pi}\sin(nx)\,dx= \int_{-\pi}^{\pi}\sin(nx)=0,$$

because $\sin$ is an odd function. No induction needed. Proving the equation for $\cos$ looks difficult without using $u$-substitution in some form (you wrote in a comment that you weren't supposed to use substitution).

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Draw the graphs. The function $x\mapsto \sin(nx)$ has $n$ complete periods on the interval $[0,2\pi]$. Each hump above the graph is geometrically congruent to its neighbor, which is below the $x$-axis. As a result, the signed area under the graph. $$\int_0^{2\pi}\sin(nx)\,dx = 0,$$ is zero.

You can see a similar phenomenon with the cosine function.

This can be turned into a calculaion easily

$$\int_0^{2\pi} \sin(nx)\, dx = \sum_{k=1}^n \int_{2(k-1)\pi/n}^{2k\pi/n} \sin(nx)\,dx = \int_0^{2\pi/n} \sin(x)\,dx = 0$$

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  • $\begingroup$ I understand this intuitively, but I'm not sure I could turn this in as a proof. $\endgroup$
    – Jackson
    Oct 22, 2013 at 22:22
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$$\displaystyle z=a+ib=\int_{0}^{2\pi}\cos{nx}+i\int_{0}^{2\pi}\sin{nx}\\=\int_{0}^{2\pi}e^{inx}dx=\frac{1}{in}(e^{i2\pi}-1)=1-1=0$$ $$\displaystyle z=a+ib=0\Rightarrow a=b=0$$

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