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This integral popped up when is was trying to solve this. I don't know if it's possible to get a general solution for

$$I = \int_0^1 x^n\arcsin^2(x) \, dx$$

where $n\in\mathbb{N}$. WolframAlpha is able to solve the integral for $n=1,2,3$, but then the computation time runs out. One can substitute $u = \arcsin(x)$ and $du = \frac{1}{\sqrt{1-x^2}} \, dx$

$$I = \int_0^{\frac{\pi}{2}}x^2\sin^n(x)\sqrt{1-\sin^2(x)} \, dx = \int_0^{\frac{\pi}{2}} x^2\sin^n(x)\cos(x) \, dx$$

WolframAlpha uses some trigonometric identities to solve the cases when $n=1,2,3$, but is there a way to solve for all $n\in\mathbb{N}$?

Casimir Rönnlöf
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    Two times by parts first to get rid of $x^2$. – Alexey Burdin Aug 18 '20 at 12:42
  • When you do the substitution, you need to update the limits of integration. The upper limit should change to $\frac{\pi}{2}$ – Moko19 Aug 18 '20 at 13:10
  • @Moko19 Thanks for noting, I forgot to change them – Casimir Rönnlöf Aug 18 '20 at 13:11
  • Elaborating the comment by @AlexeyBurdin, one IBP takes you to $\int_0^{\pi/2}x\sin^m x,dx$, for which there's a recurrence (obtained by further IBP like discussed here), and a formula with a summation, obtained the way it's done in the linked stuff. – metamorphy Aug 18 '20 at 13:41
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    Another way: using the series $2\arcsin^2\left(\frac x2\right)=\sum_{k\ge1}\frac{x^{2k}}{k^2\binom{2k}k}$ yields $\frac12\sum_{k\ge1}\frac{4^k}{(2k+n+1)k^2\binom{2k}k}$. One should be able to transform this series into a Generalized Hypergeometric Function matching the result given below. – mrtaurho Aug 18 '20 at 13:46
  • Sometimes Mathematica is much faster if you perform the indefinite integration first, then substitute extrema by hands. – Vincenzo Tibullo Aug 18 '20 at 15:06

2 Answers2

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I know it may not help you with how to evaluate, but Mathematica gives the solution $$ \frac{2 \, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{n}{2}+2;1\right)}{(n+ 1) (n+2)}+\frac{\pi ^2}{4 (n+1)}-\frac{\pi ^{3/2} \Gamma \left(\frac{n}{2}+1\right)}{(n+1)^2 \Gamma \left(\frac{n}{2}+\frac{1}{2}\right)} $$ which also appears to work for at least some fractional $n$. $\;_3F_2$ uses the notation of a generalised hypergeometric function. The right most term is related to the Mellin transform of $\arcsin^2(x)$.

Mathematica's solution is probably reached by using the representation of $\arcsin(x)$ as a Meijer-G function and solving a general form for the integral of a pair of Meijer-G functions. Finally, converting the result back down to a hypergeometric function. This is a common algorithm for symbolically solving integrals in general, but it's hard to say for sure, as your integral is also convolved with a Heaviside step function.

It's more likely that you can write your integral as $\mathcal{M}[\Theta(1-x) \arcsin^2(x)]$, i.e. the Mellin transform of the product of $\Theta(1-x)$ and $\arcsin^2(x)$, which have Meijer-G representations $$ \Theta(1-x) = \text{MeijerG}(\{\{\},\{1\}\},\{\{0\},\{\}\},x) $$ and $$ \arcsin^2(x) = -\frac{1}{2} \sqrt{\pi } \text{MeijerG}\left(\{\{1,1,1\},\{\}\},\left\{\{1\},\left\{0,\frac{1}{2}\right\}\right\},i x,\frac{1}{2}\right) $$ and use the equation $$ \int_0^{\infty} G_{p,q}^{\,m,n} \!\left( \left. \begin{matrix} \mathbf{a_p} \\ \mathbf{b_q} \end{matrix} \; \right| \, \eta x \right) G_{\sigma, \tau}^{\,\mu, \nu} \!\left( \left. \begin{matrix} \mathbf{c_{\sigma}} \\ \mathbf{d_\tau} \end{matrix} \; \right| \, \omega x \right) dx = \frac{1}{\eta} \; G_{q + \sigma ,\, p + \tau}^{\,n + \mu ,\, m + \nu} \!\left( \left. \begin{matrix} - b_1, \dots, - b_m, \mathbf{c_{\sigma}}, - b_{m+1}, \dots, - b_q \\ - a_1, \dots, -a_n, \mathbf{d_\tau} , - a_{n+1}, \dots, - a_p \end{matrix} \; \right| \, \frac{\omega}{\eta} \right) $$ or similar, so the computer is a very helpful tool, especially for breaking the result apart in terms of hypergeometric identities.

Benedict W. J. Irwin
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An alternative solution, avoiding special functions.

Sometimes an indefinite integral can be obtained if one make an ansatz about the solution, depending on some unknown parameters, then by differentiating, the correct value of the parameters can be obtained.

Assume that for even $n=2m$ the solution has the form $$ \int x^{2m}\arcsin^2(x)dx=-2xP_m(x^2)+2\sqrt{1-x^2}Q_m(x^2)\arcsin(x)+\frac{x^{2m+1}}{2m+1}\arcsin^2(x)+C $$ where $P_m,Q_m$ are polynomial of degree $m.$ Then, by differentiating we have the identity $$ -2P_m(x^2)-4x^2P'_m(x^2)-\frac{2x}{\sqrt{1-x^2}}Q_m(x^2)\arcsin(x)+4x\sqrt{1-x^2}Q'_m(x^2)\arcsin(x)+\\+2Q_m(x^2)+x^{2m}\arcsin^2(x)+\frac{x^{2m+1}}{2m+1}\frac{1}{\sqrt{1-x^2}}2\arcsin(x). $$ All terms must vanish, except for $x^{2m}\arcsin^2(x)$, so, separating the terms that contain $\arcsin(x)$ from the others, and with the position $t=x^2,$ we have the two first order linear differential equations: $$ 2(1-t)Q'_m-Q_m+\frac{t^m}{2m+1}=0\\ 2tP'_m+P_m-Q_m=0 $$ of which we don't need, and don't want, the general solutions, that contain square roots, but only the unique particular polynomial solutions. Once found these solutions, it is easy to see that the value of the definite integral is $$ \int_0^1 x^{2m}\arcsin^2(x)dx=\frac{1}{2m+1}\left(\frac{\pi}{2}\right)^2-2P_m(1). $$

In a similar way, for odd $n=2m+1$, we suppose $$ \int x^{2m+1}\arcsin^2(x)dx=-x^2P_m(x^2)+2x\sqrt{1-x^2}Q_m(x^2)\arcsin(x)+\left(\frac{x^{2m+2}}{2m+2}-k\right)\arcsin^2(x)+C $$ and going directly to the differential equation obtained, they are $$ t(1-t)Q'_m+(1-2t)Q_m+\frac{t^{m+1}}{2m+2}-k=0,\\ tP'_m+P_m-Q_m=0 $$ (from the first of these we also get $k=Q_m(0)$).
Again, we look for the polynomial solution, and once found, we have $$ \int_0^1 x^{2m+1}\arcsin^2(x)dx=\left(\frac{1}{2m+2}-Q_m(0)\right)\left(\frac{\pi}{2}\right)^2-P_m(1). $$

Vincenzo Tibullo
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