Evaluation of $\int \frac{x\sin(\sin x)}{x+5} \ dx$ How do we find $$\int  \frac{x\sin(\sin x)}{x+5} \ dx\ ,$$ is there any way to take that $\sin x$ out from parent $\sin(\cdot)$ ? 
 A: 
How do we find $\displaystyle\int\frac{x\sin(\sin x)}{x+5}dx$ ?

We don't. See Liouville's theorem and the Risch algorithm for more details. More to the point, 
even the more humble looking $\displaystyle\int_0^\tfrac\pi2\sin(\sin x)~dx=\frac\pi2H_0(1)$ requires the presence of Struve  functions in its expression.
A: Let $u=x+5$ ,
Then $x=u-5$
$dx=du$
$\therefore\int\dfrac{x\sin(\sin x)}{x+5}dx$
$=\int\dfrac{(u-5)\sin(\sin(u-5))}{u}du$
$=\int\sin(\sin(u-5))~du-\int\dfrac{5\sin(\sin(u-5))}{u}du$
For $\int\sin(\sin(u-5))~du$ ,
$\int\sin(\sin(u-5))~du$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}(u-5)}{(2n+1)!}du$
$=-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}(u-5)}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(1-\cos^2(u-5))^n}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}C_k^n(-1)^k\cos^{2k}(u-5)}{(2n+1)!}d(\cos(u-5))$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k}(u-5)}{(2n+1)!k!(n-k)!}d(\cos(u-5))$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}(u-5)}{(2n+1)!k!(n-k)!(2k+1)}+C$
