how to integrate ∫x^2/√((1-x^2))
Let x = sinϴ
dx = cosϴ dϴ
x² = sin²ϴ
ϴ = arcsin x
ʃx² dx / √(1-x²)
= ʃsin²ϴ * cosϴ dϴ / √(1-sin²ϴ)
= ʃsin²ϴ * cosϴ dϴ / √(cos²ϴ)
= ʃsin²ϴ * cosϴ dϴ / cosϴ
= ʃsin²ϴ dϴ
= ½ ʃ 1 - cos2ϴ dϴ
= ½ϴ - ¼sin(2ϴ) + C
Now sub back in ϴ = arcsin x
½arcsin x - ¼sin(2arcsinx) + C
sin(2arcsinx) can be futher simplified as follows:
Remember arcsinx = ϴ, so we are actually finding sin2ϴ.
If sinϴ = x, then cosϴ =√(1-x²)
sin2ϴ = 2sinϴcosϴ = 2x√(1-x²)
multiply it by ¼ is ½x√(1-x²). So you may see an answer that looks like:
½arcsin x - ½x√(1-x²) + C
let x=sin(theta) dx=cos(theta)d(theta) substitute in the integral to obtain:
Integral(sin^2(theta)d(theta)) replace sin^2(theta) by (1-cos(2(theta))/2 and integrate
(1/2)(theta-sin(2theta)/2)+C now go back to x
(1/2)(sin^-1(x)-xsqrt(1-x^2))+C
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Verified answer
Let x = sinϴ
dx = cosϴ dϴ
x² = sin²ϴ
ϴ = arcsin x
ʃx² dx / √(1-x²)
= ʃsin²ϴ * cosϴ dϴ / √(1-sin²ϴ)
= ʃsin²ϴ * cosϴ dϴ / √(cos²ϴ)
= ʃsin²ϴ * cosϴ dϴ / cosϴ
= ʃsin²ϴ dϴ
= ½ ʃ 1 - cos2ϴ dϴ
= ½ϴ - ¼sin(2ϴ) + C
Now sub back in ϴ = arcsin x
½arcsin x - ¼sin(2arcsinx) + C
sin(2arcsinx) can be futher simplified as follows:
Remember arcsinx = ϴ, so we are actually finding sin2ϴ.
If sinϴ = x, then cosϴ =√(1-x²)
sin2ϴ = 2sinϴcosϴ = 2x√(1-x²)
multiply it by ¼ is ½x√(1-x²). So you may see an answer that looks like:
½arcsin x - ½x√(1-x²) + C
let x=sin(theta) dx=cos(theta)d(theta) substitute in the integral to obtain:
Integral(sin^2(theta)d(theta)) replace sin^2(theta) by (1-cos(2(theta))/2 and integrate
(1/2)(theta-sin(2theta)/2)+C now go back to x
(1/2)(sin^-1(x)-xsqrt(1-x^2))+C