Conclude integral is zero without calculation I have the following two integrals:
$$\int_0^{2\pi} \frac{\cos{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
and I know both are zero. I can argue why the second one is so:
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
Therefore,
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}+\int_{0}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
With the variable change $\phi=-\theta$ I get $\sin{\theta}d\theta=\sin{\phi}d\phi$ because sine is an odd function (also, the denominator does not change), so the first term becomes
$$\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{\pi}^{0} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}}=-\int_{0}^{\pi} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}},$$
and it cancels the second term so the sine integral is zero.
Is there a similar procedure for the cosine function? Since it is an even funcion the argument can't be exactly the same. Any help will be appreciated!

Hint: Use the substitution $\phi = \pi/2 - \theta$.
