# How to evaluate $\int \sin(x)\arcsin(x)dx$

I need to evaluate following integral

$$\int \sin(x)\arcsin(x) \ dx$$

• Maybe write the second term as $\arcsin{x}$ to avoid confusion :) Commented Mar 14, 2015 at 13:47
• I would be surprised is this integral has a closed form in terms of standard functions. Commented Mar 14, 2015 at 13:53
• This integral seems unsolvable Commented Mar 14, 2015 at 13:54
• One problem might be that $\sin:\mathbb{R}\mapsto [-1,1]$ whereas $\arcsin:[-1,1]\mapsto [-\pi/2,\pi/2]$. Commented Mar 14, 2015 at 13:55
• i tried $t = \sin x,$ needed the $\int \sin(\sin t) \, dt$ and that is as hard to find as the original one.
– abel
Commented Mar 14, 2015 at 13:56

I shall not repeat that there is probably no closed form for the integral. But Taylor expansion could help; using the standard formulas for $\sin(x)$ and $\arcsin(x)$, one can obtain for the product $$\sin(x)\arcsin(x)=x^2+\frac{x^6}{18}+\frac{x^8}{30}+\frac{2669 x^{10}}{113400}+\frac{601 x^{12}}{34020}+\frac{726587 x^{14}}{52390800}+O\left(x^{16}\right)$$ and, integrating, $$\int\sin(x)\arcsin(x)\, dx=\frac{x^3}{3}+\frac{x^7}{126}+\frac{x^9}{270}+\frac{2669 x^{11}}{1247400}+\frac{601 x^{13}}{442260}+\frac{726587 x^{15}}{785862000}+O\left(x^{17}\right)$$ which, taking into account Demosthene's remark about the maximum range for integration, will converge quite rapidly.

• claude, where do you get these series so fast?
– abel
Commented Mar 14, 2015 at 14:22
• @abel. Did you notice my age ? For all that time, I am in love with Taylor series and making products of so simple expression is just a piece of my best cakes. Cheers :-) Commented Mar 14, 2015 at 14:24
• @ClaudeLeibovici, Sir, I like that comment. Very positive Commented Mar 14, 2015 at 14:30
• i have noticed your age. i hope when i get there i will still be doing and enjoying math as much as you do now and will do for a long time.
– abel
Commented Mar 14, 2015 at 14:30
• @abel. Thank you ! I am sure you will do even much more. Good luck ! Commented Mar 14, 2015 at 14:33

The integrand does not possess an elementary anti-derivative. However, as far as definite integrals are concerned, we have the following results in terms of the special Bessel and Struve functions:

\begin{align} &\int_0^1\sin x\cdot\arcsin x~dx~=~\frac\pi2~\Big[J_0(1)-\cos(1)\Big], \\ &\int_0^1\sin x\cdot\arccos x~dx~=~\frac\pi2~\Big[1-J_0(1)\Big], \\ &\int_0^1\cos x\cdot\arcsin x~dx~=~\frac\pi2~\Big[\sin(1)-H_0(1)\Big], \\ &\int_0^1\cos x\cdot\arccos x~dx~=~\frac\pi2~H_0(1). \end{align}

From the comments, there seems to be "no closed form in terms of standard functions" (to quote Simon S). Also, you have to be careful that the domains of the two functions are not the same: $\sin(x)$ takes values from $\mathbb{R}$, but the domain of $\arcsin(x)$ is restricted to $[-1,1]$.

However, one can perform numerical integration without too much difficulties. In particular, integrating on the full (real) range of $\arcsin(x)$ yields the interesting result:

$$\int_{-1}^1\sin(x)\arcsin(x)\ dx\simeq \dfrac{\sqrt{2}}{2}-5.7\times 10^{-4}$$