I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that

$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$

I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!

  • $\begingroup$ Are you sure you don't mean to expand the integral? The integrand is the part we know how to expand from the binomial theorem, not with this integral. $\endgroup$
    – J.G.
    Nov 1, 2019 at 17:01
  • $\begingroup$ term by term integration $\endgroup$
    – Narasimham
    Nov 2, 2019 at 10:40

8 Answers 8


If I was doing this I would start with one of the very common series like $\sin(x)$, $e^x$, etc. and then use substitution. The one that fits best here in my opinion is: $$ \sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \dotsc $$ Through substitution, we can obtain: $$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + \dotsc $$ Then by integration: $$ \arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc $$

  • 13
    $\begingroup$ What is this substitution? $\endgroup$
    – Solon
    Mar 14, 2018 at 10:11
  • 2
    $\begingroup$ I think the substitution is $x=\dfrac{y^2}{1-y^2}$. $\endgroup$
    – Jimmy
    Sep 22, 2018 at 6:29
  • $\begingroup$ What substitution are you talking about @Daniel Fischer $\endgroup$
    – 324
    Nov 2, 2019 at 20:06
  • 2
    $\begingroup$ @324 I'm not sure what noahklein had in mind. Possibly to replace $x$ with $-x^2$ in the series for $\sqrt{1+x}$, and then use $\frac{1}{\sqrt{1-x^2}} = \frac{1}{1-x^2}\sqrt{1-x^2} = (1 + x^2 + x^4 + \ldots)\sqrt{1-x^2}$. But I'd rather go with Norbert's or Sasha's answer. $\endgroup$ Nov 2, 2019 at 20:19
  • 1
    $\begingroup$ Another approach would be to differentiate the series for $\sqrt{1+x}$, then substitute $-x^2$ for $x$. $\endgroup$ May 2, 2023 at 0:33

Hint: Use the following binomial series $$ (1+z)^\alpha=\sum\limits_{k=0}^{+\infty}\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}z^k $$ with $z=-t^2$, $\alpha=-1/2$. Then integrate.


Change variables $t = x u$: $$ \arcsin(x) = \int_0^1 \frac{ x}{(1- x^2 u^2)^{1/2}} \mathrm{d} u $$ Now using Taylor series expansion of the integrand: $$ \frac{ x}{(1- x^2 u^2)^{1/2}} = \sum_{n=0}^\infty \frac{1}{2^{2n}}\binom{2n}{n} x^{2n+1} u^{2n} $$ Integrating term-wise: $$ \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} $$


As has been mentioned in other answers, the series for $\frac1{\sqrt{1-x^2}}$ is most easily found by substituting $x^2$ into the series for $\frac1{\sqrt{1-x}}$. But for fun we can also derive it directly by differentiation.

To find $\frac{\mathrm d^n}{\mathrm dx^n}\frac1{\sqrt{1-x^2}}$ at $x=0$, note that any factors of $x$ in the numerator produced by differentiating the denominator must be differentiated by some later differentiation for the term to contribute at $x=0$. Thus the number of contributions is the number of ways to group the $n$ differential operators into pairs, with the first member of each pair being applied to the numerator and the second member being applied to the factor $x$ produced by the first. This number is non-zero only for even $n=2k$, and in that case it is $\frac{(2k)!}{2^kk!}$. Each such pair accumulates another factor $1\cdot3\cdot\cdots\cdot(2k-1)=\frac{(2k)!}{2^kk!}$ from the exponents in the denominator. Thus the value of the $n$-th derivative at $x=0$ is $\frac{(2k)!^2}{4^k(k!)^2}$, so the Maclaurin series is

$$ \frac1{\sqrt{1-x^2}}=\sum_{k=0}^\infty\frac1{(2k)!}\frac{(2k)!^2}{4^kk!}x^{2k}=\sum_{k=0}^\infty\frac{\binom{2k}k}{4^k}x^{2k}\;. $$

Then integrating yields

$$ \arcsin x=\sum_{k=0}^\infty\frac{\binom{2k}k}{4^k(2k+1)}x^{2k+1}\;. $$


The following Python + SymPy script:

from sympy import *

# symbolic variable
x = Symbol('x')

# compute and print Maclaurin series
print series(asin(x),x,0,11,'+')

produces the following output:

x + x**3/6 + 3*x**5/40 + 5*x**7/112 + 35*x**9/1152 + O(x**11)

Therefore, assuming that the implementation is correct, the answer to your question is:

$$\displaystyle\arcsin(x) \approx x + \frac{1}{6} x^3 + \frac{3}{40} x^5 + \frac{5}{112} x^7 + \frac{35}{1152} x^9$$


If all you want is the terms of $\arcsin x$ out to $x^5$, it’s also easy enough to invert (“revert”) the series $x-x^3/6+x^5/120$ by hand. You’re looking for a series $A(x)=x + ax^3+bx^5$ so that $\sin(A(x))=x$, but you’re willing to do it modulo terms of degree six or more. Just working modulo $(x^4)$, you see immediately that $a=+1/6$, so the only work is to find $b$. When you calculate $\sin(A(x))$ out to degree five, always ignoring any term in $x^n$ for $n>5$, you get $x+(b-1/12+1/120)x^5$, and you want that last part to be zero, so $b=3/40$.


Considering the binomial expansion \begin{align} \frac 1{\sqrt{1-u}}&=1+\frac12 u+\frac12\cdot\frac32\frac{u^2}{2!}+\frac12\cdot\frac32\cdot\frac52\frac{u^3}{3!}++\frac12\cdot\frac32\cdot\frac52\cdot\frac 72\frac{u^4}{4!}+\dotsm \\ &=1+\frac12 u+\frac{1\cdot3}{2^2\,2!}u^2 +\frac{1\cdot3\cdot 5}{2^3\,3!}u^3 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}u^4\dotsm , \end{align} and substituting $u=x^2$, then integrating term by term, you obtain \begin{align} \arcsin x&= x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2^2\,2!}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2^3\,3!}\frac{x^7}7 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}\frac{x^9}9+\dotsm \\[1ex] &=x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2\cdot 4}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2\cdot4\cdot 6}\frac{x^7}7 +\frac{1\cdot3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\frac{x^9}9+\dotsm \\[1ex] &=\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}. \end{align}


Recall the series $$(1-x)^{-a}=\,_1F_0(a;;x)=\sum_{n\ge0}\frac{(a)_n}{n!}x^n$$ where $(s)_n=\frac{\Gamma(s+n)}{\Gamma(s)}=(s+n-1)(s+n-2)\cdots s$. Hence $$\begin{align} \arcsin z&=\int_0^z \frac{dt}{\sqrt{1-t^2}}\\ &=\int_0^z\sum_{n\ge0}\frac{(1/2)_n}{n!}t^{2n}dt\\ &=\sum_{n\ge0}\frac{(1/2)_n}{n!}\int_0^z t^{2n}dt\\ &=\sum_{n\ge0}\frac{(1/2)_n}{n!}\frac{z^{2n+1}}{2n+1}. \end{align}$$


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