I am trying to show that,
\begin{align*} \frac{1-\cos x}{x} = \int_{0}^{\pi/2}J_1(x\cos\theta)\,\mathrm{d}\theta \end{align*}
I did the following but cannot figure out how to continue.
\begin{align*} \int_{0}^{\pi/2}J_1(x\cos\theta)\,\mathrm{d}\theta &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2(n+1)(n!)^24^n} \int_{0}^{\pi/2}\cos^{2n+1}\theta\,d\theta\cr &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2(n+1)(n!)^24^n}\cdot \frac{(n!)^24^n}{(2n+1)!}\cr &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2(n+1)(2n+1)!} \end{align*}
Observe that $2(n+1)(2\,n+1)!=(2\,n+2)!$. Then $$ \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2(n+1)(2n+1)!}=\frac1x\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2\,n+2)!}. $$