1
$\begingroup$

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.

$\endgroup$

closed as off-topic by Jean-Claude Arbaut, Claude Leibovici, Watson, user228113, user91500 Sep 24 '16 at 12:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jean-Claude Arbaut, Claude Leibovici, Watson, Community, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please tell us what you've tried. $\endgroup$ – user37238 Sep 27 '13 at 9:46
3
$\begingroup$

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 $$\int_0^{2\pi k} \alpha \sin\left(\frac{x}{k}\right) + \beta \cos\left(\frac{x}{k}\right) dx = 0 \qquad \forall\ \alpha,\beta,k \in \mathbb R$$

$\endgroup$
  • $\begingroup$ You mean, the "computation" was hidden in "$2\pi$-periodic functions with mean $0$". (Not really a problem of course, just pointing it out.) $\endgroup$ – Lord_Farin Sep 27 '13 at 9:49
  • $\begingroup$ @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 ;-) $\endgroup$ – AlexR Sep 27 '13 at 9:51
  • $\begingroup$ Indeed; nice addition in your edit. +1. $\endgroup$ – Lord_Farin Sep 27 '13 at 10:01
1
$\begingroup$

$$ \int \sin\theta + \frac{\cos\theta}{2} d\theta=\cos\theta-\frac{\sin\theta}{2}$$ So $$ \int_0^{2\pi}\sin\theta + \frac{\cos\theta}{2} d\theta=1-1-0+0=0$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Lord_Farin Sep 27 '13 at 9:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.