Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to prove that the integral: $$ \int_0^{2\pi} f(x)\cos(x)\, dx $$ is positive. It has continuous first and second derivatives and such that $f''(x)>0$ for $0<x<2π$. I know I need to use the integration by parts and the fundamental theorem of calculus. But I'm not sure how such a thing can be proved.

share|cite|improve this question

By integration by parts, $$ \int_0^{2\pi}f(x)\cos xdx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin xdx=-\int_0^{2\pi}f'(x)\sin xdx. $$ By integration by parts once more, $$ -\int_0^{2\pi}f'(x)\sin xdx=f'(x)\cos(x)|_0^{2\pi}-\int_0^{2\pi}f''(x)\cos(x)dx\\ =f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos(x)dx=\int_0^{2\pi}f''(x)(1-\cos(x))dx\ge0 $$ where the last equality uses the fundamental theorem of calculus.

share|cite|improve this answer
I prefer this proof because of the last equality. – marty cohen Sep 5 '14 at 0:46

We use integration by parts:

$$\begin{align*}\int_0^{2 \pi} f(x) \cos x dx&=f(x) \sin x |_0^{2 \pi} - \int_{0}^{2 \pi} f'(x) \sin x dx\\&=-\int_0^{2 \pi} f'(x) \sin x dx\\&=f'(x) \cos x|_0^{2 \pi}+ \int_0^{2 \pi} f''(x) \cos x dx\\&=f'(2 \pi)-f'(0)-\int_0^{2 \pi} f''(x) \cos x\end{align*}$$

So,now,it remains to show that:

$$f'(2 \pi)-f'(0) \geq \int_0^{2 \pi} f''(x) \cos x dx$$

We know that $\cos x \leq 1 \Rightarrow f''(x) \cos x \leq f''(x)$


$$\int_0^{2 \pi} f''(x) \cos x dx \leq \int_0^{2 \pi} f''(x) dx=f'(2\pi)-f'(0)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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