Proof that $\int_0^{2\pi}\sin nx\,dx=\int_0^{2\pi}\cos nx\,dx=0$ 
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.

I do not know if you can use $e^{ix}=\cos x+i\sin x$ but here is one solution:
$$
\int_0^{2\pi}e^{inx}\,dx=\frac 1{in}e^{inx}\left|_0^{2\pi}\right.=0
$$
