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Math Help - Trig substitution calc 2

  1. #1
    Newbie
    Joined
    Jul 2010
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    13

    Trig substitution calc 2

    Dear MHF experts (btw, you guys rock!! lol),

    \displaystyle \int \frac{x^2}{\sqrt{25-x^2}}\,dx

    (someone else's calculations for this, last step is correct, i took the derivative and it equals the original integral)

    let sin(θ) = x/5 ; 5 cos(θ) dθ = dx
    = 25 ∫ sin²(θ) dθ
    = 25/2 ∫dθ - 25/2 ∫cos(2θ) dθ ............. sin²(θ) = (1/2) [1 - cos(2θ)]
    = 25/2 θ - 25/2 sin(θ)cos(θ) + c
    = 25/2 arcsin(x/5) - x/2 √(25 - x²) + c

    ----------------------------------------------------------------------------
    I got to the step in bold no problem. I got the first term in the last step right; however, I cannot seem get from the second term in bold to the second term in the last step. I only need help getting from the bold underlined to the last underlined term xD

    In the last step that's underlined, I can substitute (x/5)=sinθ ... but cannot figure out what to substitute for cosθ
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  2. #2
    Junior Member
    Joined
    Sep 2010
    Posts
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    sin \theta = x/5; cos ^2 \theta + sin ^2 \theta =1; cos \theta = \sqrt{1-x^2/25};  cos \theta = 1/5 \sqrt{25-x ^2}<br />
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