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x = 2cos\theta y = 3sin2\theta[br]\[br]\dfrac{dy}{d\theta} = 6cos2\theta, \dfrac{dx}{d\theta} = -2sin\theta[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6cos2\theta}{-2sin\theta}[br]\[br]\Rightarrow Using\ cos2\theta = 1 - 2sin^2 \theta[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6 - 12sin^2 \theta}{-2sin\theta}[br]\[br]\Rightarrow \dfrac{dy}{dx} = 6sin\theta - 3cosec\theta
m_{tangent}\ at\ \theta=\dfrac{\pi}{6} \Rightarrow sin\dfrac{\pi}{6} = \dfrac{1}{2}[br]\[br]m_{tangent} = -3 \Rightarrow m_{normal} = \dfrac{1}{3}
y^2 = 9sin^2 2\theta[br]\[br]Using\ sin2\theta = 2cos\theta sin\theta[br]\[br]\Rightarrow y^2 = 36cos^2 \theta sin^2 \theta[br]\[br]cos^2 \theta = \dfrac{x^2}{4}\ and\ sin^2 \theta = 1 - cos^2 \theta[br]\[br]\Rightarrow y^2 = \dfrac{9}{4}x^2 (4-x^2)
9x^2 - 6xy + 4y^2 = 3[br]\[br]implicit\ differentiation[br]\[br]18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}[br]\[br]\Stationary\ at \dfrac{dy}{dx} =0[br]\[br]3x-y = 0 \Rightarrow y = 3x[br]\[br]\Sub\ this\ into\ original\ equation[br]\[br]27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1
\dfrac{5x-6}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}[br]\[br]5x-6 = A(x-3) + Bx[br]\[br]Let x = 3 \Rightarrow 9 = 3B \Rightarrow B = 3[br]\[br]Let x = 0 \Rightarrow -6 = -3A \Rightarrow A = 2[br]\[br]\dfrac{5x-6}{x(x-3)} = \dfrac{2}{x} + \dfrac{3}{x-3}[br]
4x^3 +5x -2 = (2x+1)(2x^2 +px + q) + r[br]\[br]Using\ factor\ theorem,\ do\ f(\dfrac{-1}{2})[br]\[br]4(\dfrac{-1}{2})^3 + 5\dfrac{-1}{2} -2 = -5,\ therefore\ r = -5[br]\ [br]Now,\ 4x^3 +5x +3 = (2x+1)(2x^2 +px + q),\ so\ by\ inspection\ p = -1 and q = 3
Now\ divide\ the\ function\ previously\ found[br]\[br]\Rightarrow \dfrac{2}{3}x^3 -\dfrac{1}{2}x^2 + 3x -\dfrac{5}{2} ln(2x+1) + c
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