Integral $\int x\sin ax \cos x\, dx$ I'm looking for a quick and elegant  way to calculate the integral:
$$\int x\sin ax \cos x\, dx$$
It's doable by using $$\cos x=\frac{e^{ix}+e^{-ix}}2, ~~\sin ax=\frac{e^{iax}-e^{-iax}}{2i},$$
and then integrating by parts, but I find that solution too long and brute-force. Can you solve it alternatively?
Edit: 
$$\sin ax\cos x=\frac{\sin (a+1)x+\sin (a-1)x}2$$
does the trick.
 A: Use Werner formula:
$$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$
Now integrate by parts $$\int x\sin(px)\ dx=x\int\sin(px)\ dx-\int\left(\dfrac{dx}{dx}\int\sin(px)\ dx\right)dx$$
OR as $$\dfrac{d(x\cos(px))}{dx}=\cos(px)-px\sin(px)$$
Integrate both sides, $$x\cos(px)+K=\int\cos(px)\ dx-p\int x\sin(px)\ dx$$
A: $$I(x;a,b):=-\int \cos ax\cos bx\,dx=-\frac{\sin((a+b)x)}{2(a+b)}-\frac{\sin((a-b)x)}{2(a-b)}$$
and
$$\int x\sin ax\cos x\,dx=\frac\partial{\partial a}I(x;a,1).$$
A: Just with some (mild) trigonometry to linearise :
$$\sin ax\,\cos x=\frac12\bigl(\sin(a+1)x+\sin(a-1)x\bigr),$$
then integration by parts to remove the $x$ factor.
A: If you don't have trig identities at your fingertips, note that
$$\int x\sin ax\cos x\,dx=-{dI\over da}$$
where
$$I(a)=\int\cos ax\cos x\,dx$$
Integration by parts twice tells us
$$\begin{align}
I(a)
&=\cos ax\sin x+a\int\sin ax\sin x\,dx\\
&=\cos ax\sin x-a\sin ax\cos x+a^2\int\cos ax\cos x\,dx\\
&=\cos ax\sin x-a\sin ax\cos x+a^2I(a)
\end{align}$$
hence
$$I(a)={\cos ax\sin x-a\sin ax\cos x\over1-a^2}$$
and thus
$$\begin{align}
\int x\sin ax\cos x\,dx
&={(1-a^2)(-x\cos ax\sin x-\sin ax\cos x-ax\cos ax\cos x)+2a(\cos ax\sin x-a\sin ax\cos x)\over(1-a^2)^2}\\
&={(a^2x-x+2a)\cos ax\sin x-(a^2+1)\sin ax\cos x+a(a^2-1)\cos ax\cos x\over(1-a^2)^2}
\end{align}$$
Remark: At the end of all this, one should add the obligatory "constant" of integration $C$, which in this case is a function, $C(a)$. Among other things, the constant of integration can help reconcile two different-looking answers obtained by two different approaches to evaluating the integral. (Its main purpose is to ensure you don't have points taken off by a picky instructor....)
A: $$2\int x\sin ax\cos x dx=\int x\sin(a-1)x dx+\int x\sin(a+1)xdx =I_1+I_2$$ by the formula: $2\sin A\cos B=\sin(A+B)+\sin(A-B)$
Consider $I_1:$ set $u=(a-1)x$, then $du=(a-1)dx$, then 
$$\int x\sin(a-1)x dx=\int \frac{u\sin u}{(a-1)^2}du=\frac{1}{(a-1)^2}\int u\sin u du=\underbrace{\cdots}_{\text{integration by parts}}=\frac{1}{(a-1)^2}\big(\sin u- u\cos u\big)$$
Consider now $I_2:$ set $v=(a+1)x$, then $dv=(a+1)dx$, then 
$$\int x\sin(a+1)x dx=\int \frac{v\sin v}{(a+1)^2}dv=\frac{1}{(a+1)^2}\int v\sin v dv=\underbrace{\cdots}_{\text{integration by parts}}=\frac{1}{(a+1)^2}\big(\sin v- v\cos v\big)$$
Summing up $I_1+I_2$ we get 
$$ \frac{1}{(a-1)^2}\big(\sin u- u\cos u\big)+\frac{1}{(a+1)^2}\big(\sin v- v\cos v\big)$$
Finally: $$\int x\sin ax\cos x dx=\frac12\Big[\frac{1}{(a-1)^2}\big(\sin (a-1)x- (a-1)x\cos (a-1)x\big)+ \frac{1}{(a+1)^2}\big(\sin (a+1)x- (a+1)x\cos (a+1)x\big)\Big]$$
