Wallis's integral and Lambert's function . Inspired by the Wallis's integral I propose to calculate the following integral :


$$\int_{0}^{e}\sin(\operatorname{W(x)})^ndx$$
    Where we have the Lambert's function and $n\geq1$ a natural number 


I can calculate the first :


$$\int_{0}^{e}\sin(\operatorname{W(x)})dx$$


The antiderivative is :
$$\int_{}^{}\sin(\operatorname{W(x)})dx=\frac{(x (\sin(\operatorname{W(x)}) + W(x) (\sin(\operatorname{W(x)}) - \cos(\operatorname{W(x)}))))}{2 \operatorname{W(x)}} + \operatorname{constant}$$
And the result is :
$$e(\sin(1)-\frac{\cos(1)}{2})$$
I can continue like this to calculate the antiderivative but I don't see the link between each integral ($n$ to $n+1$) . I have tried integration by part but I failed .I just know that when the number is odd or even we have all the number odd or even in the cosine or the sine before the power.This remark make a link with Wallis's integral where we find all the odd and even number before the power.  
Any ideas ?
Thanks a lot . 
 A: Note that $\sin(t)^n$ can be written as a linear combination of $\sin(k t)$ (if $n$ is odd) or $\cos(k t)$ (if $n$ is even), and you can get antiderivatives for $\sin(k W(x))$ and $\cos(k W(x))$.
$$\int \!\sin \left( k{\rm W} \left(x\right) \right) \,{\rm d}x= \left( 
{\frac {x}{{k}^{2}+1}}+2\,{\frac {{k}^{2}x}{ \left( {k}^{2}+1 \right) 
^{2}{\rm W} \left(x\right)}} \right) \sin \left( k{\rm W} \left(x
\right) \right) + \left( -{\frac {kx}{{k}^{2}+1}}-{\frac {k \left( {k}
^{2}-1 \right) x}{ \left( {k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)
}} \right) \cos \left( k{\rm W} \left(x\right) \right) 
$$
$$ \int \!\cos \left( k{\rm W} \left(x\right) \right) \,{\rm d}x= \left( 
{\frac {kx}{{k}^{2}+1}}+{\frac {k \left( {k}^{2}-1 \right) x}{ \left( 
{k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)}} \right) \sin \left( k
{\rm W} \left(x\right) \right) + \left( {\frac {x}{{k}^{2}+1}}+2\,{
\frac {{k}^{2}x}{ \left( {k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)}
} \right) \cos \left( k{\rm W} \left(x\right) \right)
$$
Alternatively, express in terms of complex exponentials:
$$ \int \exp(ik {\rm W}(x))\; dx = {\frac {{{\rm e}^{ \left( ik+1 \right) {\rm W} \left(x\right)}}
 \left( (ik+1) {\rm W} \left(x\right)+ik \right) }{
 \left( ik+1 \right) ^{2}}}
$$
A: Starting from @Robert Israel's answer, considering
$$I_k=\int \sin \big[ k\,W (x) \big] \,dx$$
$$k\,W (x)=t \implies x=\frac{t}{k}\, e^{\frac{t}{k}}\implies dx=\frac{k+t}{k^2}e^{\frac{t}{k}} $$
$$I_k=\frac 1 {k^2}\int (k+t)\,e^{\frac{t}{k}}  \sin (t)\,dt$$
$$I_k=\frac{e^{\frac{t}{k}} }{k \left(k^2+1\right)^2}\,\Big[\left(2 k^3+(k^2+1) t\right) \sin (t)-k \left(k(k^2-1)+(k^2+1) t\right) \cos (t) \Big]$$
Similarly
$$J_k=\int \exp \big[i\, k\,W (x) \big] \,dx=-\frac{i\, e^{\left(\frac{1}{k}+i\right) t}\, \left(k^2+(k-i) t\right)}{k (k-i)^2}$$
