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.
Best Answer
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*}