evaluate the following integral $$
\int J_0(x)\sin x~{\rm d}x
$$
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?
 A: 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)
$$
\begin{aligned}
D\bigl(x J_1(x)\cos(x)\bigr)
&=J_1(x)\cos(x)+xJ_1'(x)\cos(x)-xJ_1(x)\sin (x)\\
&=J_1(x)\cos(x)+x\frac{1}{2}(J_0(x)-J_2(x))\cos(x)-x J_1\sin(x)\\
&=J_1(x)\cos(x)+x\frac{1}{2}\bigl(J_0(x)-(-J_0(x)+\frac{2}{x}J_1(x))\bigr)\cos(x)-x J_1\sin(x)\\
&=x J_0(x)\cos(x)-xJ_1(x)\sin(x).
\end{aligned}
$$
A: I know that I answered this question after so long time. But the main fact is I have a solution in another way and that's why I'm  here. And I think this new way will help people a lot to find out the solution. 
..................................................................................................................................................................
\begin{equation}I=\int J_0(x) \sin(x) dx\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}-\int J_0(x) e^{-ix}\right)dx\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}~dx+\int J_0(-y) e^{iy}~dy\right)\\
\left[\text{Second part change the integration variable to $y=-x$}\right]\\
\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}~dx+\int J_0(y) e^{iy}~dy\right)\\
\left[\text{By the parity of the Bessel function}\right]\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))+e^{iy}y(J_0(y)−iJ_1(y))\right\}+c\\
\left[\text{where $~c~$ is a constant}\right]\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))-e^{-ix}x(J_0(-x)−iJ_1(-x))\right\}+c\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))-e^{-ix}x(J_0(x)+iJ_1(x))\right\}+c\\
\left[\text{By the parity of the Bessel function}\right]\\
=\dfrac x{2i}\left\{J_0(x)(e^{ix}-e^{-ix})-iJ_1(x)(e^{ix}+e^{-ix})\right\}+c\\
=x[\sin x · J_0(x) − \cos x · J_1(x)]+c\\
\\\end{equation}
Hence $$\boxed{\int J_0(x) \sin(x) dx=x\left\{\sin x · J_0(x) − \cos x · J_1(x)\right\}+c}$$where $~c~$ is a constant.
..................................................................................................................................................................


*

*Parity of the Bessel function : $~J_n(-x)=(-1)^nJ_n(x)~$

*$$\int J_0(x) e^{ix}~dx=x~e^{ix}\left\{J_0(x)−iJ_1(x)\right\}$$
