Intrigued by this brilliant answer from Ron Gordon, I was attempting to find the Maclaurin series for $$f(x)=\frac{\arcsin x}{\sqrt{1-x^2}}=g(x)G(x)$$

with $g(x)=\frac{1}{\sqrt{1-x^2}}$ and $G(x)$ its primitive. So I attempted to multipy series, which yielded this:

$$f(x)=\sum_{n=0}^{\infty}x^{2n+1} (-1)^n\sum_{k=1}^{n}\frac{1}{k+1} { -\frac{1}{2}\choose n-k}{ -\frac{1}{2}\choose k},$$

which I'm unable to simplify further. How to proceed? Or is this approach doomed?

  • $\begingroup$ Maybe try expanding $\sqrt{1 - x^2}$ via the binomial theorem, and then dividing. $\endgroup$
    – MattyZ
    Nov 2, 2013 at 23:24
  • 1
    $\begingroup$ Since $f(x)=\frac12(\arcsin^2x)'$, perhaps you should try express this new function in Taylor-Maclaurin series, and then derive it with regard to x. $\endgroup$
    – Lucian
    Nov 23, 2013 at 17:59
  • $\begingroup$ @Bitrex No. not really it is too symbolic! $\endgroup$
    – Z Ahmed
    Jul 8, 2020 at 7:20
  • $\begingroup$ @Lucian Thanks, I am going to try this and work on. $\endgroup$
    – Z Ahmed
    Jul 8, 2020 at 7:21

4 Answers 4


Note that $$\int_0^1\frac{dt}{1-x^2+x^2t^2}=\frac{1}{x\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{\arcsin(x)}{x\sqrt{1-x^2}}$$ so we can write $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n\,dt\right)x^{2n+1}.$$ But $$\int_0^1(1-t^2)^n\,dt=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}t^{2k}\,dt=\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}.$$ Hence, $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}.$$ Also see here for proof of $\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$.

  • 1
    $\begingroup$ Hey, can you explain the transition from the first line to the second ? I really looked hard at this, yet frustratingly enough I can't see how you did that. $ \frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n\,dt\right)x^{2n+1}. $ Where does the integral appear from and how does it relate to the previous one ? $\endgroup$
    – Victor
    Sep 27, 2015 at 3:39
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    $\begingroup$ @Victor Note that $\sum_{n=0}^\infty(1-t^2)^nx^{2n}$ is a geometric series, so in fact,$$\sum_{n=0}^\infty(1-t^2)^nx^{2n}=\frac1{1-(1-t^2)x^2}=\frac1{1-x^2+x^2t^2}$$ Now, take a look once again! $\endgroup$
    – user91500
    Sep 27, 2015 at 11:37
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    $\begingroup$ That is so slick ! I know this might be really simple for you, but how did you come up with this? I basically mean the first integral and the geometric series part. $\endgroup$
    – Victor
    Sep 27, 2015 at 15:12
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    $\begingroup$ @Victor Oh no, I think the problem is that you like to see from the left to right, while you can also see from the right to left! $\endgroup$
    – user91500
    Sep 28, 2015 at 9:32
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    $\begingroup$ In fact, you want to find the Maclaurin series for $\frac{\arcsin(x)}{\sqrt{1-x^2}}$, among the all things you may know about this function, one thing can be the following equality $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}\arctan\frac{x}{\sqrt{1-x^2}}$$ So if you try to find an integral representation for $\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$, this can be useful and really useful when the integrand has also a geometric series representation! $\endgroup$
    – user91500
    Sep 28, 2015 at 9:33

Another but similar proof which does not need to use the summation formula above is this one. Start with defining

$$I(t)= \frac{1}{\sqrt{1-x^2}}\arctan{\frac{x\sin{t}}{\sqrt{1-x^2}}}$$

Then by the Fundamental theorem of calculus

$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=I\left(\frac{\pi}{2}\right)-I(0)=\int_{0}^{\pi/2} \frac{\partial I}{\partial t}\mathrm{d}t=\int_{0}^{\pi/2}\frac{x\cos t}{1-x^2\cos^2 t }\mathrm{d}t$$


$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n+1}\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$$

Denote $J_n:=\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$, by per partes we have

$$J_n = \int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t = 2n\int_0^{\pi/2}\cos^{2n-1}\sin^2 t\!\,\mathrm{d}t=2n\left(J_{n-1}-J_{n}\right)$$

So $$J_n = \frac{2n}{2n+1}J_{n-1} =\frac{2n}{2n+1}\frac{2n-2}{2n-1}J_{n-2}=\dots = \frac{(2n)!!}{(2n+1)!!}J_0=\frac{(2n)!!}{(2n+1)!!}$$

since $J_0 = \int_0^{\pi/2}\cos\! t\,\mathrm{d}t =1$. Over all we get desired result


Note: Similar integral would have been also...

$$\int_{0}^{\pi/2}\frac{\mathrm{d}t}{1-x\sin t}$$


Another proof is using that $f(x)={{\arcsin x}\over {\sqrt{1-x^2}}}$ satisfies $(1-x^2)f'=xf+1$, which can be easily verified.


It is clear that \begin{equation*} \frac{\arcsin x}{\sqrt{1-x^2}\,}=\frac12\bigl[(\arcsin x)^2\bigr]'. \end{equation*} Accordingly, it is sufficient to find out the power series expansion of $(\arcsin x)^2$ around $x=0$. The fact is \begin{equation*} (\arcsin x)^2=2!\sum_{n=0}^{\infty} [(2n)!!]^2 \frac{x^{2n+2}}{(2n+2)!}, \quad |x|<1. \end{equation*} A more general conclusion than the above power series expansion is \begin{equation}\label{arcsin-series-expansion-unify} \biggl(\frac{\arcsin x}{x}\biggr)^{k} =1+\sum_{m=1}^{\infty} (-1)^m\frac{Q(k,2m)}{\binom{k+2m}{k}}\frac{(2x)^{2m}}{(2m)!}, \end{equation} where \begin{equation}\label{Q(m-k)-sum-dfn} Q(k,m)=\sum_{\ell=0}^{m} \binom{k+\ell-1}{k-1} s(k+m-1,k+\ell-1)\biggl(\frac{k+m-2}{2}\biggr)^{\ell} \end{equation} for $k\in\mathbb{N}$ and $m\ge2$.


  1. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.
  2. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Applicable Analysis and Discrete Mathematics 16 (2022), no. 2, 427--466; available online at https://doi.org/10.2298/AADM210401017G.
  3. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Mathematics 6 (2021), no. 7, 7494--7517; available online at https://doi.org/10.3934/math.2021438.
  4. F. Qi, Explicit formulas for partial Bell polynomials, Maclaurin's series expansions of real powers of inverse (hyperbolic) cosine and sine, and series representations of powers of Pi, Research Square (2021), available online at https://doi.org/10.21203/rs.3.rs-959177/v3.
  5. https://math.stackexchange.com/a/4661247.
  6. https://math.stackexchange.com/a/4657809.
  7. https://math.stackexchange.com/a/4379986.

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