# Evaluate $\int\sin(\sin x)~dx$

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.

• The Jacobi-Anger expansion gives $\sin(\sin(x))$ in terms of an infinite sine series weighted by Bessel functions: en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion Commented Jan 26, 2013 at 16:38
• @Bitrex: very interesting. Something new to me. Thank you! Commented Jan 26, 2013 at 16:41
• It looks so simple but it looks difficult also. Commented Jan 26, 2013 at 17:07
• @NancyR: yeah. That's true. Commented Jan 26, 2013 at 17:54
• Commented Mar 4, 2013 at 15:05

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.

• OK. Thanks for your suggestion. (+1) Commented Jan 26, 2013 at 16:46
• Do you mean $du$?
– leo
Commented Jan 26, 2013 at 17:10
• Oops, yes of course Commented Jan 26, 2013 at 17:32
• Out of curiosity if you deifne x to be a number like you would when looking for an answer say $\frac {3^{1/2}}{2}$ can u just draw it on a triangle then look for the angle on the triangle then use that angle and opp/hyp of that angle to find the side of that triangle then integrate that constant over your bounds? Commented May 10, 2013 at 13:16
• @Faust7: not entirely sure what you are asking. When doing trig substitutions, it is wise to use a right triangle as your guide in relating sines, cosines, etc. But I do not understand "integrating the constant over the bounds." $x$ is a variable and is expressed in terms of another variable, $u$. That's it. Commented May 10, 2013 at 13:21

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 $$\boxed{\int \sin (\sin x) d x =\sum_{n=0}^{\infty} \sum_{k=0}^n {n\choose k} \frac{(-1)^{n+k+1} \cos ^{2 k+1} x}{(2 k+1)(2n+1)!}+C}$$

• The expression with the sums seems to be more complicated than the original expression. Commented Jul 3, 2023 at 8:52
• Yes, it is not satisfactory. Any alternative?
– Lai
Commented Jul 3, 2023 at 10:54