anonymous
  • anonymous
how do you integrate y* sqrt(6+4y-4y^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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RadEn
  • RadEn
first, 6+4y-4y^2 = -(4y^2 - 4y -6) = -((2y - 1)^2 - 7)) u = 2y - 1 ------> y = (u+1)/2 du = 2 dy du/2 = dy see the integral above becomes : int ( (u+1)/2) * sqrt(-(u^2 - 7)) 1/2 du = 1/4 int (u+1) * sqrt(7-u^2) du looks we have to use trigono subt u= sqrt(7)sinθ --------> u^2 = 7sin^2 θ, then sqrt(7-u^2) = sqrt(7-7sin^2 θ) = sqrt(7(1-sin^2 θ)) = sqrt(7cos^2 θ) = sqrt(7)cosθ du = sqrt(7)cosθ dθ 1/4 int (sqrt(7)sinθ + 1) * sqrt(7)cosθ * sqrt(7)cosθ dθ = 1/4 int (sqrt(7)sinθ + 1) * 7 cos^2 θ dθ = 7/4 int (sqrt(7)sinθcos^2 θ + cos^2 θ) dθ split the integral above be 2 parts : 7/4 int (sqrt(7)sinθcos^2 θ) dθ + 7/4 intcos^2 θ dθ
RadEn
  • RadEn
the 1st part is using u-sub, and the 2nd part is change the equation by using identity trigono : cos^2 θ = (1 + cos2θ)/2 = 1/2 + 1/2 cos2θ

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