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In the process of tackling a question I've ended up having to evaluate the integral $$ I = \displaystyle \int_0^1 \dfrac{4 \sin(2 \pi t) \sin (4 \pi t) + 2 \cos (2 \pi t ) \cos (4 \pi t)}{\sin^2 (2 \pi t) + \cos ^2 (4 \pi t ) } $$

By checking online integral calculators I can see that this integral should vanish but I'm not sure how to go about proving it. In order to make the domain of integration symmetric I tried the substitution $ t \mapsto t - \frac{1}{2} $ leading to the similar expression $$ I = - \displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{4 \sin(2 \pi t) \sin (4 \pi t) + 2 \cos (2 \pi t ) \cos (4 \pi t)}{\sin^2 (2 \pi t) + \cos ^2 (4 \pi t ) }. $$

From here I attempted to use the substitution $ t \mapsto - t $ but I recovered the same expression for $I$ that was attained previously.

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    $\begingroup$ Hint: f(-x)=f(x) $\endgroup$ – user411437 Nov 6 '18 at 18:56
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    $\begingroup$ the denominator is 1 $\endgroup$ – clathratus Nov 6 '18 at 20:43
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Note the fraction that you got in the second expression is the same as the one in the first expression. The only differences between the two are the limits and the negative sign.

Now look at each trigonometric function. $1$ is a common period, so the expression to integrate has a period of $1$. We can then integrate any interval of length $1$ and we get the same result. Simplified notation: $$I=\int_0^1=\int_{-1/2}^{1/2}$$ But you already have $$I=-\int_{-1/2}^{1/2}=-I$$ so $I=0$

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  • $\begingroup$ Slick approach, I was trying to do it directly via substitution and never thought to utilize the periodicity of the integrand :) $\endgroup$ – backstrapp Nov 6 '18 at 21:05
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Consider that: $$\sin(4\pi t)=2\sin(2\pi t)\cos(2\pi t)$$ And: $$\cos(4\pi t)=1-2\sin^2(2\pi t)$$ Then: $$I=\int_0^1\frac{\cos(2\pi t)[4\sin^2(2\pi t)+2]}{1-3\sin^2(2\pi t)+4\sin^4(2\pi t)}dt$$ Now let $u=\sin(2\pi t)$ so $du=\cos(2\pi t)2\pi dt$ for the integration limit we observe that $u(0)=u(1)=0$ then: $$I=\frac{1}{2\pi}\int_0^0\frac{4u^2+2}{1-3u^2+4u^4}du=0$$

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  • $\begingroup$ Is there not an issue with the substitution not being injective over the given domain? $\endgroup$ – backstrapp Nov 6 '18 at 21:02
  • $\begingroup$ The chage of variable works; try to see that if you calculat your intergral from t=0 to t=1/3 then I=2/3 in both ways $\endgroup$ – Alessio Bocci Nov 6 '18 at 23:58

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