# Prove that :$\frac{1}{100\pi}>\int_{100 \pi}^{200\pi}\frac{\cos(x)}{x}>0$

Using integration by parts, prove that $$0<\int_{100\pi}^{200\pi}\frac{\cos(x)}{x}<\frac{1}{100\pi}.$$

Using integration by parts prove that $\frac{1}{100\pi}>\int_{100 \pi}^{200\pi}\frac{\cos(x)}{x}>0$.

Could anyone give me a help with this problem? I have tried using integration by parts but I don't get what integrating by parts achieves. Intuitively I can see why the integral must be greater than 0 as $\cos(100\pi) = \cos(200\pi) = 1$ and $\frac{1}{x}$ is a decreasing function, so the various areas above and below the $x$ axis and $\frac{\cos(x)}{x}$ would cancel to something positive, but this isn't using integration by parts.

We integrate by parts. letting $u=1/x$ and $dv=\cos x\,dx$. So $du=-\frac{1}{x^2}\,dx$, and we can take $v=\sin x$. Our integral is equal to $$\left. \frac{1}{x}\sin x\right|_{100\pi}^{200\pi}+\int_{100\pi}^{200\pi} \frac{\sin x}{x^2}\,dx.$$ The first part vanishes, and the remaining integrand has absolute value less than $\frac{1}{(100\pi)^2}$ except at the very beginning. The interval has length $100\pi$, so the integral has absolute value less than $100\pi\cdot \frac{1}{(100\pi)^2}$.
It only remains to show that the integral is positive. To do this, it is enough to show that the integral from $2k\pi$ to $2k\pi+2\pi$ is positive. We are looking at $$\int_{2k\pi}^{2k\pi+\pi}\frac{\sin x}{x^2}\,dx+\int_{2k\pi+\pi}^{2k\pi+2\pi}\frac{\sin x}{x^2}\,dx.\tag{1}$$ For the second integral, make the change of variable $t=x-\pi$. After a little manipulation we find rhat the sum of the integrals in (1) is $$\int_{2k\pi}^{2k\pi+\pi}\sin x\left(\frac{1}{x^2}-\frac{1}{(x+\pi)^2}\right)\,dx,$$ which is clearly positive.