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. $\endgroup$
    – Bitrex
    Nov 2, 2013 at 23:24
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    $\begingroup$ 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

2 Answers 2


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 $\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$.

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    $\begingroup$ 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...

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


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