I was skimming the virtual pages here and noticed a limit that made me wonder the following
question: is there any nice way to evaluate the indefinite integral below?
$$\int\sin(\sin x)~dx$$ Perhaps one way might use Taylor expansion. Thanks for any hint, suggestion.
Maybe you could do something like substitute $u=\sin{x}$ and get
$$\int du \: \frac{\sin{u}}{\sqrt{1-u^2}}$$
You could Taylor expand the denominator and be in a position to integrate even moments of $\sin{u}$ and see if the resulting series is useful.
For the maclaurin series of $\sin x$ , $\sin x=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$
$\therefore\int\sin(\sin x)~dx=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx$
Now for $\int\sin^{2n+1}x~dx$ , where $n$ is any non-negative integer,
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}x}{(2n+1)!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k+1}n!\cos^{2k+1}x}{k!(n-k)!(2n+1)!(2k+1)}+C$
Using Euler identity:$e^{xi}=\cos x+i\sin x$, we have $$ \begin{aligned} \int \sin (\sin x) d x & =\operatorname{Im} \int e^{i \sin x} d x \\ & =\operatorname{Im} \sum_{n=0}^{\infty} \int \frac{(i \sin x)^n}{n !} d x \\ & =\operatorname{Im} \sum_{n=0}^{\infty} \int \frac{i^{2 n+1} \sin ^{2 n+1} x}{(2 n+1) !} d x \\ & =\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1) !} \int \sin ^{2 n+1} x d x \end{aligned} $$ Putting $u=\cos x$ transforms the integral into $$ \begin{aligned} \int \sin ^{2 n+1} x d x & =-\int\left(1-u^2\right)^n d u =-\sum_{k=0}^n\left(\begin{array}{c} n \\ k \end{array}\right) \frac{(-1)^k u^{2 k+1}}{2 k+1} \end{aligned} $$ Now we can conclude that $$ \boxed{\int \sin (\sin x) d x =\sum_{n=0}^{\infty} \sum_{k=0}^n {n\choose k} \frac{(-1)^{n+k+1} \cos ^{2 k+1} x}{(2 k+1)(2n+1)!}+C} $$
