$$ \int J_0(x)\sin x~{\rm d}x $$
Where $J_0$ is Bessel function of first kind of order $0$
This what I tried
$$ \int J_0(x)\sin x~{\rm d}x= -J_0(x) \cos x - \int J_0'(x)\cos x~{\rm d}x $$
$$ J_0'(x)=-J_1(x) $$
$$ \int J_0(x)\sin x ~{\rm}x= -J_0(x) \cos x -(J_1(x)\sin x - \int J_1'(x)\sin x~{\rm d}x) $$
$$ \int J_0(x)\sin x~{\rm d}x=-J_0(x) \cos x - J_1(x) \sin x +\left(\int J_0(x)\sin x~{\rm d}x + \int(\sin x/ x) J_1(x)~{\rm d}x\right) $$
But this won't help to evaluate it, is there any other method?
We integrate by parts, $$ \begin{aligned} \int J_0(x)\sin(x)\,dx &=xJ_0(x)\sin(x)-\int x\bigl(-J_1(x)\sin(x)+J_0(x)\cos(x)\bigr)\,dx\\ &=xJ_0(x)\sin(x)-\int D\bigl(x J_1(x)\cos(x)\bigr)\,dx\\ &=xJ_0(x)\sin(x)-x J_1(x)\cos(x)+C. \end{aligned} $$
Clarification
In the second step, we used the well-known recurrence relations for Bessel functions, $$ \begin{aligned} \frac{2}{x}J_1(x)=J_0(x)+J_2(x),\quad\text{and}\quad 2J_1'(x)=J_0(x)-J_2(x) \end{aligned} $$ to get (the calculation here goes backwards) $$ \begin{aligned} D\bigl(x J_1(x)\cos(x)\bigr) &=J_1(x)\cos(x)+xJ_1'(x)\cos(x)-xJ_1(x)\sin (x)\\ &=J_1(x)\cos(x)+x\frac{1}{2}(J_0(x)-J_2(x))\cos(x)-x J_1\sin(x)\\ &=J_1(x)\cos(x)+x\frac{1}{2}\bigl(J_0(x)-(-J_0(x)+\frac{2}{x}J_1(x))\bigr)\cos(x)-x J_1\sin(x)\\ &=x J_0(x)\cos(x)-xJ_1(x)\sin(x). \end{aligned} $$