Evaluating $\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$ 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
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\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.

A: i will use series expansion and find a recurrence relation.  first define
$$ J_k = \int_0^x t^{2k}\sin t \,dt, k = 0, 1, \dots $$
we have $J_0 = \int_0^x \sin \, dt = 1 - \cos x.$  now,
$\begin{align}
J_k &= \int_0^x t^{2k} \sin t \, dt = -t^{2k}\cos t|_0^x + \int_0^x 2k t^{2k-1} \cos t\, dt\\
&= 2kt^{2k-1}\sin t-t^{2k}\cos t|_0^x - 2k(2k-1)\int_0^x t^{2k-2} \sin t\, dt\\
&= 2kx^{2k-1}\sin x - x^{2k}\cos x -2k(2k-1)J_{k-1} = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1}
\end{align}$
the recurrence relation is $$J_k = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1}\,  , \, J_0 = 1 - \cos x\tag 1 $$
we can compute  $$J_1 = x(2\sin x - x\cos x) -2(1-\cos x),\\ J_2 = x^3(4\sin x - x \cos x )-12x(2\sin x - x \cos x)+24(1-\cos x), \dots$$
$$\int_0^x \frac{\sin t}{1 + t^2}\, dt = \int_0^x \sin t\left(1 -t^2 + t^4 + \dots\right) = J_0- J_1 + J_2 + \dots, \text{ for } -1  < x < 1.$$
A: The relevant power series expansions in the radii of convergence (0,x) are 
   $$\sin (x) = \sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}$$
     $$ \frac{1}{1+{t^2}} = 1+x^2+x^4+\ldots= \sum_{j=0}^{\infty} {x^{2j}}$$  
So now we multiply the power series and integrate term by term: 
     $$\frac{\sin(x)}{1+{t^2}}$$ = $ \sum_{j=0}^{\infty} {x^{2j}}\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}$ = $ \sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} x^{4j+1}$
We obtained the last part by adding together like coeffecients as follows: 1 = $-1^{2}$ and $ \sum_{j=0}^{\infty} {x^{2j}}\cdot x^{2j+1}$= $ \sum_{j=0}^{\infty} {x^{4j+1}}$. So now we simply integrate term by term,which yields 
  $$\int_0^x \dfrac{\sin(t)}{1+t^2}dt =\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} \frac{x^{4j+2}}{(4j+2)!}$$
A: $\int_0^x\dfrac{\sin(t)}{1+t^2}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n+1}}{(2n+1)!(t^2+1)}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^n(t^2+1-1)^n}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(t^2+1)^k}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{\sinh1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\left[\dfrac{\sinh1\ln(t^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^k}{2(2n+1)!k!k(n-k)!}\right]_0^x$
$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!((x^2+1)^k-1)}{2(2n+1)!k!k(n-k)!}$
