------------
Welcome to March Calculus Madness!
------------
∫x^2 / √(1 - x^2) dx
Substitute: z = arcsin x
sin z = x
sin^2 z = x^2
and dz = 1/√(1 - x^2) dx
Then:
∫x^2 / √(1 - x^2) dx
= ∫ sin^2 z dz
= ∫ 1/2 - 1/2 ∙ cos(2∙z) dz
(by trigonometric identity of sin^2 z = 1/2 - 1/2 ∙ cos(2∙z))
= z/2 - 1/4 ∙ sin(2∙z) + C
= z/2 - 1/2 ∙ sin z ∙ cos z + C
(by trigonometric identity of sin(2∙z) = 2 ∙ cos z ∙ sin z)
= 1/2 ∙ arcsin x - 1/2 ∙ sin(arcsin x) ∙ cos(arcsin x) + C
= 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - x^2) + C
(sin(arcsin x) = x, cos(arcsin x) = √(1 - x^2)
Summary:
∫x^2 / √(1 - x^2) dx = 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - x^2) + C
Eddie
All original content copyright, © 2011-2022. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.