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Welcome to March Calculus Madness!
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∫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
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