<s>[INST] 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
 [/INST] 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  $$J_1 = x(2\sin x - x\cos x) -2(1-\cos x),\\ J_2 = x^3(4\sin x - x \cos x )-12x(2\sin x - x \cos x)+24(1-\cos x), \dots$$
$$\int_0^x \frac{\sin t}{1 + t^2}\, dt = \int_0^x \sin t\left(1 -t^2 + t^4 + \dots\right) = J_0- J_1 + J_2 + \dots, \text{ for } -1  < x < 1.$$
</s>