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.
 A: 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*}
A: I do not know if you can use $e^{ix}=\cos x+i\sin x$ but here is one solution:
$$
\int_0^{2\pi}e^{inx}\,dx=\frac 1{in}e^{inx}\left|_0^{2\pi}\right.=0
$$
A: 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 $\int_{\pi/n}^{2\pi/n}\cos(nx)\,\mathrm{d}x$ ?
A: 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 $\cos$?
A: 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).
A: 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
$$\int_0^{2\pi} \sin(nx)\, dx = \sum_{k=1}^n \int_{2(k-1)\pi/n}^{2k\pi/n}
\sin(nx)\,dx = \int_0^{2\pi/n} \sin(x)\,dx = 0$$
A: $$\displaystyle z=a+ib=\int_{0}^{2\pi}\cos{nx}+i\int_{0}^{2\pi}\sin{nx}\\=\int_{0}^{2\pi}e^{inx}dx=\frac{1}{in}(e^{i2\pi}-1)=1-1=0$$
$$\displaystyle z=a+ib=0\Rightarrow  a=b=0$$
