# Showing integral vanishes using symmetry

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.

• Hint: f(-x)=f(x) – user411437 Nov 6 '18 at 18:56
• the denominator is 1 – clathratus Nov 6 '18 at 20:43

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