Find $\lim_{x\to 0}\frac{\int_0^1\arctan(t+\sin x)-\arctan tdt}{\arctan x}$ I want to find 
$$
\lim_{x \to 0}\left\{{1 \over \arctan\left(x\right)}
\int_{0}^{1}
\left[\arctan\left(t + \sin\left(x\right)\right)-\arctan\left(t\right)\right]\,{\rm d}t\right\}
$$
Here is what I did:
By mean value theorem, we can find 
$$\frac{1}{1+\xi^2}=\frac{\arctan(t+\sin x)-\arctan t}{\sin x}$$
which means
$$\arctan(t+\sin x)-\arctan t=\frac{\sin x}{1+\xi^2}$$
Then the limit can be convert to:
$$\lim_{x\to 0}\frac{\int_0^1\frac{\sin x}{1+\xi^2}d\xi}{\arctan x}
=\lim_{x\to 0}\frac{\sin x}{\arctan x}[\arctan\xi]_0^1=\frac{\pi}{4}$$
I am not sure if I am doing this right. If not， what is the right way to do this?
 A: Notice $$
\int_0^1 \tan^{-1}(t+\tau) - \tan^{-1} t  dt
= \left(\int_{\tau}^{1+\tau} - \int_0^1 \right) \tan^{-1} t dt
= \left(\int_1^{1+\tau} - \int_{0}^{\tau}\right)\tan^{-1} t dt
$$
By fundamental theorem of calculus.
$$\lim_{x\to 0}\frac{\int_0^1 \tan^{-1}(t+\sin x) - \tan^{-1} t  dt}{\sin x}
= \lim_{\tau\to 0}\frac{\int_0^1 \tan^{-1}(t+\tau) - \tan^{-1} t  dt}{\tau}\\
= \frac{d}{d\tau}\left[\left(\int_1^{1+\tau} - \int_{0}^{\tau}\right)\tan^{-1} t dt\right]_{\tau=0}
= \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4}
$$
Since $\;\;\displaystyle \lim_{x\to 0}\frac{\sin x}{\tan^{-1} x} = 1,\;\;$ the limit we seek is also $\frac{\pi}{4}$.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#88f}{\large%
\lim_{x \to 0}\braces{{1 \over \arctan\pars{x}}\int_{0}^{1}
\bracks{\arctan\pars{t + \sin\pars{x}} - \arctan\pars{t}}
\,{\rm d}t}}
\\[3mm]&=
\lim_{x \to 0}\bracks{{\sin\pars{x} \over \arctan\pars{x}}\int_{0}^{1}
{\arctan\pars{t + \sin\pars{x}} - \arctan\pars{t} \over \sin\pars{x}}
\,{\rm d}t}
\\[3mm]&=
\lim_{x \to 0}\bracks{\int_{0}^{1}
{\arctan\pars{t + x} - \arctan\pars{t} \over x}\,{\rm d}t}
=
\int_{0}^{1}\totald{\arctan\pars{t}}{t}\,{\rm d}t
=\arctan\pars{1} - \arctan\pars{0}
\\[3mm]&=\color{#88f}{\Large{\pi \over 4}}
\end{align}
A: I think you need to go beyond the mean value theorem to Taylor theorem, and write $$\arctan(t + \sin x) = \arctan t + \sin(x)/(1+t^2) + o(\sin^2(x)).$$ Then, you get the same answer as yours, but you don't have the issue of the unclear functional behavior of your $\xi.$
