Evaluate the integral $\int_0^\pi \sin{(x \cos{t}})\cos{t}\; dt$ How to evaluate:
$\int \sin{(x \cos{t}})\cos{t}\; dt$
or:
$\int_0^\pi \sin{(x \cos{t}})\cos{t}\; dt$
 A: $\int\sin(x\cos t)\cos t~dt=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}\cos^{2n+2}t}{(2n+1)!}~dt$
For $n$ is any non-negative integer,
$\int\cos^{2n+2}t~dt=\dfrac{(2n+2)!t}{4^{n+1}((n+1)!)^2}+\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin t\cos^{2k+1}t}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}\cos^{2n+2}t}{(2n+1)!}~dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}t}{2^{2n+1}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{2n+1}(k!)^2\sin t\cos^{2k+1}t}{2^{2n-2k+1}n!(n+1)!(2k+1)!}+C$
$\therefore\int_0^\pi\sin(x\cos t)\cos t~dt$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}t}{2^{2n+1}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nx^{2n+1}(k!)^2\sin t\cos^{2k+1}t}{2^{2n-2k+1}n!(n+1)!(2k+1)!}\right]_0^\pi$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi x^{2n+1}}{2^{2n+1}n!(n+1)!}$
$=\pi J_1(x)$
Specifically for $\int_0^\pi\sin(x\cos t)\cos t~dt$ ,
$\int_0^\pi\sin(x\cos t)\cos t~dt$
$=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_\frac{\pi}{2}^\pi\sin(x\cos t)\cos t~dt$
$=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_\frac{\pi}{2}^0\sin(x\cos(\pi-t))\cos(\pi-t)~d(\pi-t)$
$=\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt+\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt$
$=2\int_0^\frac{\pi}{2}\sin(x\cos t)\cos t~dt$
$=2\int_\frac{\pi}{2}^0\sin\left(x\cos\left(\dfrac{\pi}{2}-t\right)\right)\cos\left(\dfrac{\pi}{2}-t\right)~d\left(\dfrac{\pi}{2}-t\right)$
$=2\int_0^\frac{\pi}{2}\sin(x\sin t)\sin t~dt$
$=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt-\int_0^\frac{\pi}{2}\cos(x\sin t+t)~dt$
$=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt-\int_\pi^\frac{\pi}{2}\cos(x\sin(\pi-t)+\pi-t)~d(\pi-t)$
$=\int_0^\frac{\pi}{2}\cos(x\sin t-t)~dt+\int_\frac{\pi}{2}^\pi\cos(x\sin t-t)~dt$
$=\int_0^\pi\cos(x\sin t-t)~dt$
$=\pi J_1(x)$
