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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!

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3 Answers 3

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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.

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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 $a$].

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  • $\begingroup$ By what way we can know the period of $f$ ? $\endgroup$
    – RTN96202
    Aug 25, 2020 at 8:12
  • $\begingroup$ Just replace $x$ by $x+2\pi$ and use the periodicity if $\sin $ and $\cos. @AnhTuấnTrầnNguyễn $\endgroup$ Aug 25, 2020 at 8:15
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Hint: Substitute $u=x-{2\pi}$ and consider parity

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  • $\begingroup$ How we can choose $2\pi$? $\endgroup$
    – RTN96202
    Aug 25, 2020 at 8:07
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    $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – Anand
    Aug 25, 2020 at 9:37
  • $\begingroup$ @Anand There is no reason for me to finish an homework exercise, in fact this is frowned upon on MSE in general. The user has already said in their post that they are considering integration of an odd function over $(-a,a)$. A hint is all they needed. $\endgroup$
    – jamesv
    Aug 25, 2020 at 10:42

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