Trigonometric functions expressed as definite integrals with Bessel functions Prove that 
$$\frac{\sin(x)}{x}=\int_0^\frac{\pi}{2}J_0(x\cos(\theta))\cos(\theta)\,d\theta \tag{a}$$
$$\frac{1-\cos(x)}{x}=\int_0^\frac{\pi}{2}J_1(x\cos(\theta))\,d\theta  \tag{b}$$
 Hint:
$$\int_0^\frac{\pi}{2}\cos^{2s+1}(\theta)\,d\theta  = \frac{2\cdot 4\cdot6\cdots(2s)}{1\cdot3\cdot5\cdots(2s+1)}$$
I have no  idea how to approach this problem.  Any suggestions?  I did express sine function in exponential form but then I have no clue where to go from  there so that I can end up with the integral as an answer indicated above.
 A: Since $$J_0(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{4^m\,m!^2}x^{2m}\tag{1}$$
we have:
$$\begin{eqnarray*}\int_{0}^{\pi/2}J_0(x\cos\theta)\cos\theta\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{(2m+1)!}=\frac{\sin x}{x}.\end{eqnarray*}$$
In a similar way
$$ J_1(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{2\cdot 4^m\, (m+1)\,m!^2}x^{2m+1}\tag{2}$$
gives:
$$\begin{eqnarray*}\int_{0}^{\pi/2}J_1(x\cos\theta)\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m+1}}{(2m+2)!}=\frac{1-\cos x}{x}.\end{eqnarray*}$$
$(1)$ and $(2)$ can be derived from the integral representations given in the Wikipedia page for Bessel functions.
