# Integral 1/sqrt(1-x^2)

Indefinite integral |
---|

integral 1/sqrt(1 – x^2) dx = sin^(-1)(x) + constant |

Alternate form of the integral |

-i log(sqrt(1 – x^2) + i x) + constant |

Series expansion of the integral at x=-1 |

-pi/2 + sqrt(2) sqrt(x + 1) + (x + 1)^(3/2)/(6 sqrt(2)) + (3 (x + 1)^(5/2))/(80 sqrt(2)) + (5 (x + 1)^(7/2))/(448 sqrt(2)) + (35 (x + 1)^(9/2))/(9216 sqrt(2)) + O((x + 1)^5) (Puiseux series) |

Series expansion of the integral at x=0 |

x + x^3/6 + (3 x^5)/40 + O(x^6) (Taylor series) |

Series expansion of the integral at x=1 |

1/2 (pi + (-1)^(ceiling((arg(x – 1))/(2 pi))) (-2 i sqrt(2) sqrt(x – 1) + (i (x – 1)^(3/2))/(3 sqrt(2)) – (3 i (x – 1)^(5/2))/(40 sqrt(2)) + (5 i (x – 1)^(7/2))/(224 sqrt(2)) – (35 i (x – 1)^(9/2))/(4608 sqrt(2)) + O((x – 1)^(11/2)))) |

Series expansion of the integral at x=∞ |

1/2 (2 i log(1/x) + pi – i log(4)) + i/(4 x^2) + (3 i)/(32 x^4) + O((1/x)^6) (generalized Puiseux series) |

Definite integral |

integral_(-1)^1 1/sqrt(1 – x^2) dx = pi~~3.14159 |