Find the sign of the integral $\int^{2\pi}_\pi\frac{\cos(x)}{x}dx$ (without calculating it) I am struggling to find a way to answer this question, it is under the extra reading section of my textbook so it has no method, hence me turning to here. I need to find the sign of this integral (i assume without actually calculating it).
$$I=\int^{2\pi}_\pi\frac{\cos(x)}{x}dx$$
Edit:
This is everything I have so far (I have no idea if I am on the right track).
By the Taylor series expansion for cosine, the integral becomes:
$$I=\int^{2\pi}_\pi\sum^\infty_{k=0}\frac{(-1)^kx^{2k-1}}{(2k)!}dx$$
Then continuing by integrating term-wise and getting the following:
$$I=\left.\log(x)+\sum^\infty_{k=1}\frac{(-1)^kx^{2k}}{(2k)(2k)!}\right|^{2\pi}_\pi$$
But I am struggling to group these terms in a way that reveals the sign of the integral. I am pretty sure it is negative but the log(2) is causing some issues in grouping the terms. Any help would be appreciated, whether it is th end of this method or a completely different one. Thanks!
 A: Hint: Separate the integral into two pieces, one where $\cos x$ takes positive values, the other one where it takes negative values. Can you compare the magnitudes of the two?
A: $$\begin{eqnarray*}\int_{0}^{\pi}\frac{-\cos x}{x+\pi}\,dx &=&\int_{0}^{\pi/2}\cos(x)\left[\frac{1}{2\pi-x}-\frac{1}{x+\pi}\right]\,dx\\&=&\int_{(0,\pi/2)}\underbrace{\frac{\cos x}{(2\pi -x)(\pi+x)}}_{>0}\cdot\underbrace{(2x-\pi)}_{<0}\,dx\end{eqnarray*}$$
is negative, but pretty small. Its absolute value is bounded by
$$ \int_{0}^{\pi/2}\frac{(\pi-2x)}{(2\pi-x)(\pi+x)}\,dx =\log\left(1+\frac{1}{8}\right)<\frac{1}{8}.$$

As an alternative,
$$\begin{eqnarray*} \int_{\pi}^{2\pi}\frac{\cos x}{x}\,dx &=& \log(2)-2\int_{\pi}^{2\pi}\sin^2\left(\frac{x}{2}\right)\frac{dx}{x}\\&\stackrel{\text{IBP}}{=}&\log(2)+2\log(\pi)+\int_{\pi}^{2\pi}\sin(x)\log(x)\,dx\\&=&\log(2)+2\log(\pi)-\int_{0}^{\pi}\sin(x)\log(x+\pi)\,dx\\&=&\log(2\pi^2)-\int_{0}^{\pi}\frac{\sin x}{2}\,\log\left[(x+\pi)(2\pi -x)\right]\,dx\\&=&-\int_{0}^{\pi}\frac{\sin x}{2}\,\log\left[\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\right]\,dx  \end{eqnarray*}$$
is negative and with an absolute value bounded (by the Cauchy-Schwarz inequality) by
$$ \sqrt{\int_{0}^{\pi}\left(\frac{\sin x}{2}\right)^2\,dx \int_{0}^{\pi}\log^2\left[\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\right]\,dx}=0.096338\ldots$$
