# Assistance evaluating $\int_0^{2\pi}\frac{1}{2}\left(2 \sin \theta + \cos \theta\right)d\theta$ [closed]

I need help with evaluating the following integral $$\int_0^{2\pi}\frac{1}{2}\left(2 \sin \theta + \cos \theta\right)d\theta$$ I have attempted this but I am not sure how to complete the problem.

## closed as off-topic by Jean-Claude Arbaut, Claude Leibovici, Watson, user228113, user91500Sep 24 '16 at 12:37

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• Please tell us what you've tried. – user37238 Sep 27 '13 at 9:46

It's an integration interval of length $2\pi$ over a linear combination of $2\pi$-periodic functions with mean $0$, i.e. $$\int_0^{2\pi} \sin(x) dx = \int_0^{2\pi} \cos(x) dx = 0$$ That gives $$\int_0^{2\pi} \frac{1}{2} (2\sin \theta + \cos \theta) d\theta= 0$$ without "computation".
More generally $$\int_0^{2\pi k} \alpha \sin\left(\frac{x}{k}\right) + \beta \cos\left(\frac{x}{k}\right) dx = 0 \qquad \forall\ \alpha,\beta,k \in \mathbb R$$
• You mean, the "computation" was hidden in "$2\pi$-periodic functions with mean $0$". (Not really a problem of course, just pointing it out.) – Lord_Farin Sep 27 '13 at 9:49
• @Lord_Farin yep, but for $\sin$ and $\cos$ (and more generally $\exp(ik\cdot)$), this is a very common fact. And I think it's more elegant than computing directly, as this even applies to uglier combinations ;-) – AlexR Sep 27 '13 at 9:51
$$\int \sin\theta + \frac{\cos\theta}{2} d\theta=\cos\theta-\frac{\sin\theta}{2}$$ So $$\int_0^{2\pi}\sin\theta + \frac{\cos\theta}{2} d\theta=1-1-0+0=0$$
• Someone edited your question to use \sin and \cos, which render as $\sin$ and $\cos$ (built-in shorthands looking better than $sin$ and $cos$). For further information about writing maths at this site see e.g. here, here, here and here. – Lord_Farin Sep 27 '13 at 9:53