$\int_{0}^{2\pi } \sin(\sin x+2016x)dx=?$ and $\int_{\pi }^{3\pi } \frac{\sin^{2017}(1997x)\cos^{2018}(2000x)}{1+\cos^{70}(x)+2\sin^{4}(x)}dx=?$ Problem:
Evaluate:
$$\int_{0}^{2\pi }\displaystyle \sin(\sin x+2016x)\mathrm{d}x=?$$
$$\displaystyle \int_{\pi}^{3\pi} \frac{\sin^{2017}(1997x) \cos^{2018}(2000x)}{1+\cos^{70}(x)+2\sin^{4}(x)}\mathrm{d}x=?$$

For first integral, I thought about the parity of $f(x)=\sin(\sin x+2016x)$ but the limits of the integration are not symmetrical to use: $\displaystyle \int_{-a}^{a}f(x)\mathrm{d}x=0$
Also, the integration  appears the polynomial function and the trigonometric, which makes me confused.
For $2\text{nd}$ integral, I meant to use $\displaystyle \int_{a}^{a+P}f(x)\mathrm{d}x=\int_{0}^{P}f(x)\mathrm{d}x=\int_{\frac{-P}{2}}^{\frac{P}{2}}f(x)\mathrm{d}x$, but I can't find the period of $f(x)=\frac{\sin^{2017}(1997x)\cos^{2018}(2000x)}{1+\cos^{70}(x)+2\sin^{4}(x)}$

Please help me solve these problems by following the Calculus II. Thank you!
 A: You mentioned that your main difficulty was finding $2\pi$ as a period. To do this, we can look at each term individually.
Let's look at the second integral first. All the terms look like $\sin(kx)$ or $\cos(kx)$, the period of which you may know to be $2\pi/k$. In particular, as long as $k$ is an integer, all of these are periodic with period $2\pi$ (that may not be the minimal period, say, for $\sin(1997x)$, but it is certainly a period). So, the integrand is periodic with period $2\pi$.
The first integral is a bit trickier, since the terms aren't all like $\sin(kx)$. However, note that $\sin\sin x$ should be periodic with period $2\pi$ (again, maybe there's some other smaller period, or maybe not). From this you should be able to convince yourself that a function like $\sin(\sin x+2016x)$ should be periodic with period $2\pi$; just plug in $x+2\pi$ and simplify.
A: In both cases the integrands have $2\pi$ as  a period. This implies that the integrals over any interval of length $2\pi$ are the same. Hence we can integrate from $-\pi$ to $\pi$ and conclude that both the integrals are $0$.
[Suppose $f$ is continuous and  has period $2\pi$. Then the derivative of $\int_a^{a+2\pi} f(x)dx$ w.r.t.  $a$ is $f(a+2\pi)-f(a)=0$. Hence $\int_a^{a+2\pi} f(x)dx$  is independent of $a$].
A: Hint: Substitute $u=x-{2\pi}$ and consider parity
