Showing integral vanishes using symmetry

Question

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.

Answer 1

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 $I=0$

Answer 2

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: $$I=\frac{1}{2\pi}\int_0^0\frac{4u^2+2}{1-3u^2+4u^4}du=0$$