TITLE: Evaluating $\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$
QUESTION [5 upvotes]: I cannot seem to figure out how to solve the following problem: 
$$\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$$
I have tried by using integration by parts. 
I set $u = \sin(t)$, $v = \tan^{-1}(x)$ and  $dv = \dfrac{1}{1+t^2}dt$. Thus,
$$\int_0^x \dfrac{\sin(t)}{1+t^2}dt = \left[\sin(t)\cdot \tan^{-1}(t)\right]_0^x - \int_0^x \cos(t)\cdot \tan^{-1}(t)dt$$ 
Is this correct or not? I would like to get some hints, I am currently stuck. 
Best regards

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large%
\int_{0}^{x}\frac{\sin\pars{t}}{1 + t^{2}}\,\dd t}
=\Im\int_{0}^{x}\frac{\sin\pars{t}}{t - \ic}\,\dd t
=\Im\int_{-\ic}^{x - \ic}\frac{\sin\pars{t + \ic}}{t}\,\dd t
\\[5mm]&=\Im\int_{-\ic}^{x - \ic}\frac{\sin\pars{t}\cosh\pars{1} + \cos\pars{t}\sinh\pars{1}\ic}{t}\,\dd t
\\[1cm]&=\cosh\pars{1}\bracks{\Im\int_{0}^{x - \ic}\frac{\sin\pars{t}}{t}\,\dd t
-\Im\int_{0}^{-\ic}\frac{\sin\pars{t}}{t}\,\dd t}
\\[5mm]&+\sinh\pars{1}\bracks{%
-\Re\int_{0}^{x - \ic}\frac{1 - \cos\pars{t}}{t}\,\dd t
+\Re\int_{0}^{-\ic}\frac{1 - \cos\pars{t}}{t}\,\dd t
+\Re\int_{-\ic}^{x - \ic}\frac{\dd t}{t}}
\\[1cm]&=\cosh\pars{1}\bracks{%
\Im\,{\rm Si}\pars{x - \ic} - \Im\,{\rm Si}\pars{-\ic}}
\\[5mm]&+\sinh\pars{1}\bracks{%
-\Re\,{\rm Cin}\pars{x - \ic} + \Re\,{\rm Cin}\pars{-\ic}
+\Re\ln\pars{1 + x\ic}}
\\[1cm]&=\color{#66f}{\large\cosh\pars{1}\bracks{%
\Im\,{\rm Si}\pars{x - \ic} - \Im\,{\rm Si}\pars{-\ic}}}
\\[5mm]&\color{#66f}{\large+\sinh\pars{1}\bracks{%
-\Re\,{\rm Cin}\pars{x - \ic} + \Re\,{\rm Cin}\pars{-\ic}}
+ \half\,\sinh\pars{1}\ln\pars{1 + x^{2}}}
\end{align}

$\ds{\,{\rm Si}}$ and $\ds{\,{\rm Cin}}$ are the
  Sine and Cosine Integral Functions, respectively.