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.


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.

  • 3
    $\begingroup$ I prefer this proof because of the last equality. $\endgroup$ – 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 $\cos x \leq 1 \Rightarrow f''(x) \cos x \leq f''(x)$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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