Help in Apostol Vol 1 integration review exercise 5.11.9 
Show that 
$$\displaystyle \int_0^x{\frac{\sin(t)}{t+1}\ dt} \geq 0$$
$\forall x\geq 0 $
My attempt -
By 2nd mean value theorem, 
$\displaystyle \int_0^x{\frac{\sin(t)}{t+1}\ dt} = \frac{1}{0+1} (1 - \cos(c)) + \frac{1}{x+1}(\cos(c) - \cos(x)) = 1 - \frac{x}{x+1}\cos(c) - \frac{\cos(x)}{x+1} $
But now how to show the above. 
 A: By partial integration, $$\int^x_0\frac{\sin t}{t+1}\,dt=\frac{1-\cos x}{x+1}+\int^x_0\frac{1-\cos t}{(t+1)^2}\,dt.$$ Everything on the RHS is $\ge0$ for $x\ge0.$
Remark: this device works for every integral $\int^x_0f(t)\,\sin t\,dt,$ where $f$ is positive, differentiable and monotone decreasing. The device of @robjohn wouldn't need differentiability, so it's (potentially) more general.
A: Hint: Note that
$$
\begin{align}
&\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(t)}{t+1}\,\mathrm{d}t+\int_{(2k+1)\pi}^{(2k+2)\pi}\frac{\sin(t)}{t+1}\,\mathrm{d}t\\
&=\int_0^{\pi}\frac{\sin(t)}{t+2k\pi+1}\,\mathrm{d}t-\int_0^\pi\frac{\sin(t)}{t+(2k+1)\pi+1}\,\mathrm{d}t\\
&=\pi\int_0^{\pi}\frac{\sin(t)}{(t+2k\pi+1)(t+(2k+1)\pi+1)}\,\mathrm{d}t
\end{align}
$$
That is, at the end of each interval of decrease, the integral is positive.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

This proof is valid
  $\ds{{\Huge\forall}\ x\ \in\ \mathbb{R}_{\ \geq\ 0}}$ !!! which
  "$\underline{includes}$", as a particular case, "$\ds{x \over 2\pi}$ is an
  $\ds{\underline{integer}}$" !!!.

\begin{align}
\left.\int_{0}^{x}{\sin\pars{t} \over t + 1}\,\dd t\,
\right\vert_{\ x\ \geq\ 0} & =
\int_{0}
^{\left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + \left\{x/\pars{2\pi}\right\}2\pi}
{\sin\pars{t} \over t + 1}\,\dd t
\\[5mm] & =
\int_{0}^{\left\lfloor x/\pars{2\pi}\right\rfloor 2\pi}
{\sin\pars{t} \over t + 1}\,\dd t +
\int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi}
{\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t
\\[5mm] & =
\left\lfloor{x \over 2\pi}\right\rfloor\
\overbrace{\int_{0}^{2\pi}\sin\pars{t}
\sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1}
{1 \over t  + 2\pi n + 1}\,\dd t}^{\ds{\mc{I}_{1}}}
\\[2mm] & +\
\underbrace{\int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi}
{\sin\pars{t} \over t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}
\,\dd t}
_{\ds{\equiv\ \mc{I}_{2}}}\label{1}\tag{1}
\end{align}

$\ds{\Large\mc{I}_{1} \geq 0:\ ?}$
\begin{align}
\mc{I}_{1} & =
-\int_{-\pi}^{\pi}\sin\pars{t}
\sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1}
{1 \over t  + \pars{2n + 1}\pi + 1}\,\dd t
\\[5mm] & =
-\int_{0}^{\pi}\sin\pars{t}\bracks{%
\sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1}
{1 \over t  + \pars{2n + 1}\pi + 1} -
\sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1}
{1 \over -t  + \pars{2n + 1}\pi + 1}}\,\dd t
\\[5mm] & =
2\int_{0}^{\pi}\sin\pars{t}
\sum_{n = 0}^{\left\lfloor{x/\pars{2\pi}}\right\rfloor - 1}
{t \over \bracks{\pars{2n + 1}\pi + 1}^{\,2} - t^{2}}\,\dd t\
\bbox[10px,#ffe,border:1px dotted navy]{\Large\geq 0}
\end{align}

$\ds{\Large\mc{I}_{2} \geq 0:\ ?}$


*

*It's clear that $\ds{I_{2} \geq 0}$ whenever
$\ds{\bbx{0 \leq \braces{x \over 2\pi}2\pi \leq \pi}}$.

*When $\ds{\bbx{\pi < \braces{x \over 2\pi}2\pi < 2\pi}}$:
\begin{align}
\mc{I}_{1} & =
\int_{0}^{\pi}{\sin\pars{t} \over
t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t +
\int_{\pi}^{\left\{x/\pars{2\pi}\right\}2\pi}{\sin\pars{t} \over
t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t
\\[5mm] & =
\int_{0}^{\pi}{\sin\pars{t} \over
t + \left\lfloor x/\pars{2\pi}\right\rfloor 2\pi + 1}\,\dd t -
\int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi - \pi}{\sin\pars{t} \over
t + \pars{2\left\lfloor x/\pars{2\pi}\right\rfloor + 1}\pi + 1}\,\dd t
\\[1cm] & = \overbrace{%
\int_{0}^{\left\{x/\pars{2\pi}\right\}2\pi - \pi}\sin\pars{t}\bracks{{1 \over
t + \pars{\left\lfloor x/\pars{2\pi}\right\rfloor + \color{#f00}{0}}2\pi + 1} -
{1 \over
t + \pars{\left\lfloor x/\pars{2\pi}\right\rfloor + \color{#f00}{1/2}}2\pi
+ 1}}\,\dd t}^{\large\geq\ 0}
\\[2mm] & +\ \underbrace{%
\int_{\left\{x/\pars{2\pi}\right\}2\pi - \pi}^{\pi}{\sin\pars{t} \over
t + \pars{2\left\lfloor x/\pars{2\pi}\right\rfloor + 1}\pi + 1}\,\dd t}
_{{\large \geq\ 0}.\ \mbox{See}\ \ds{1.}}\quad \bbox[#ffe,10px,border:1px dotted navy]{\Large \geq 0}
\end{align}

