How to evaluate $\int \sin(x)\arcsin(x)dx$ I need to evaluate following integral
$$\int \sin(x)\arcsin(x) \ dx$$
Can anyone please help me? Thanks.
 A: 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.
A: 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:

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

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