How to prove that $-\ln2≤\int_0^1\frac{\cos(nx)}{x+1}\,\mathrm dx≤ \ln2\\$ How can you prove that for all n, $$-\ln2≤\int_0^1\frac{\cos(nx)}{x+1}\,\mathrm dx≤ \ln2$$ 
I was thinking of using the property if $a≤x≤b,\ m≤f(x)≤M$ where $m$ and $M$ are constants, then
$$m(b-a)≤\int_a^b f(x)\,\mathrm dx≤M(b-a)$$.
But I can't prove definitely that $\dfrac{\cos(nx)}{x+1}≤ \ln2$
This is if I set $b$ as $1$, $a$ as $0$, letting $b-a=1$.
Thanks in advance for the help.
 A: $$\left|\int_0^1\frac{\cos nx}{x+1}dx\right|\leq\int_0^1\frac{1}{x+1}dx=\ln2$$
A: Observe that $-1\leq\cos(nx)\leq1$, which gives $$-\frac{1}{x+1}\leq\frac{\cos (nx)}{x+1}\leq\frac1{x+1}.$$ Now integrating from $0$ to $1$ through the inequality, we get $$-\ln(x+1)\bigg|_{x=0}^{x=1}\leq\int_0^1\frac{\cos(nx)}{x+1}\leq\ln(x+1)\bigg|_{x=0}^{x=1}.$$ Evaluating the left- and right-hand sides yields $-\ln(2)$ and $\ln(2)$, respectively.
A: You can prove much better. Let 
$f(z) = \int_{0}^{1}\frac{\cos(z x)}{x+1}\,dx$. Clearly $f$ is a differentiable function fulfilling $f(0)=\log 2, f(z)=f(-z)$, $f(z)\leq \int_{0}^{1}\frac{dx}{x+1}=\log(2)$ and by the Riemann-Lebesgue lemma $\lim_{z\to +\infty} f(z)=0$. In particular $f$ has an absolute minimum which we may show to be much larger than $-\log(2)$. Such minimum occurs at a solution of $f'(z)=0$, in particular at a positive value of $z$ such that $\int_{0}^{1}\frac{x}{x+1}\sin(zx)\,dx=0$. By integration by parts
$$f(z) = \frac{\sin(z)}{2z}+\frac{1}{z}\int_{0}^{1}\frac{\sin(zx)}{(x+1)^2}\,dx $$
and we have a curious phenomenon: the function $g(z)=\int_{0}^{1}\frac{\sin(z x)}{(x+1)^2}\,dx $ is positive for any $z>0$.
Indeed, by a fundamental property of the Laplace transform
$$ g(z) = \int_{0}^{+\infty}\underbrace{\frac{s}{e^{2s}(s^2+z^2)}}_{\geq 0}\cdot\underbrace{\left(z e^s-z\cos(z)-s\sin(z)\right)}_{\geq 0\text{ and strongly convex}}\,ds $$
hence $f(z)\geq \frac{\sin(z)}{2z}$. The absolute minimum of $\frac{\sin(z)}{2z}$ over $\mathbb{R}^+$ occurs at a point such that $\tan(z)=z$, shortly before $\frac{3\pi}{2}$. Newton's method easily locate such point in the interval $\left(\frac{3\pi}{2}-\frac{13}{18\pi},\frac{3\pi}{2}-\frac{2}{3\pi}\right)$. 
By considering the values of $\frac{\sin(z)}{2z}$ over such interval we get the improved inequality
$$ \color{red}{-\frac{1}{9}}\leq \int_{0}^{1}\frac{\cos(nx)}{x+1}\,dx \leq \log 2.$$
