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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    ∫√(1-x^2 ) dx

    Letting x = sin θ,
    dx/dθ = cos θ

    ∫√(1-x^2 ) dx
    =∫√(1-sin^2⁡θ ) cos θ dθ
    =∫cos^2 θ dθ
    =∫ (1-sin^2⁡θ) dθ
    =θ -2sinθ(-cosθ) + c
    =θ +2sinθcosθ + c
    =sin^(-1)⁡θ + 2x√(1-x^2 ) + c

    Is this correct? Please help me to check whether it is correct or not, and point out my mistake, if any. Thank you.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Integration by substitution

    Quote Originally Posted by alexander9408 View Post

    =∫√(1-sin^2⁡θ ) cos θ dθ
    =∫cos^2 θ dθ
    Hint: From here, do it by integration by parts twice.

    -Dan
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  3. #3
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    Re: Integration by substitution

    Hello, alexander9408!

    \displaystyle\int\sqrt{1-x^2}\,dx

    \text{Let }\,x = \sin\theta \quad\Rightarrow\quad dx = \cos\theta\,d\theta

    \displaystyle\text{Substitute: }\:\int\cos^2\!\theta\,d\theta

    Use the identity: . \cos^2x \:=\:\frac{1+\cos2x}{2}

    \displaystyle\text{We have: }\:\int \left(\frac{1+\cos2\theta}{2}\right)d\theta \;=\;\tfrac{1}{2}\int\left(1 + \cos2\theta\right)d\theta

    . . . . . =\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\cdot 2\sin\theta\cos\theta\right) + C

    . . . . . =\;\tfrac{1}{2}\left(\theta + \sin\theta\cos\theta\right) + C


    Back-substitute: . \tfrac{1}{2}\left(\arcsin x + x\sqrt{1-x^2}\right) + C
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