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  1. ...hence we can find perpendicular = 3 hence tan A = perpendicular/ base = -3/2 Csec is 1/ sinA and Sin A = p/h = 3/5sp Csec...

    4 Answers · Science & Mathematics · 30/08/2009

  2. ...ln|sec(t)| -2*ln|sin(t)| + C w = C*sec(t)/sin^2(t) = v' v = ∫Csec(t)/sin^2(t)) dt = ∫Ccos(t)/(1 - sin^2(t))*(sin^2(t)) dt Let z...

    1 Answers · Science & Mathematics · 30/09/2013

  3. y' - (tan x)y = 1 y ' cos x - (sin x) y = cos x y ' cos x +(cos x)' y = cos x (y cos x) ' = cos x y cos x = sin x +C y = (sin x +C)/cos x y = tan(x) +Csec(x)

    1 Answers · Science & Mathematics · 09/03/2010

  4. ... - ∫ (cos(t) sin(t) dt ucos(t) = C - (1/2)sin^2(t) u = Csec(t) - (1/2)tan(t)sin(t) Recall we substituted u = y^-1, so let...

    1 Answers · Science & Mathematics · 18/09/2008

  5. ...d/dx[ycos(x)] = cos(x) ycos(x) = sin(x) + C y = tan(x) + Csec(x) -------------- 2.) 1/√(x^2 - 3) (dy/dx) = xcos^2(y) sec^2(y) dy = x...

    1 Answers · Science & Mathematics · 25/02/2013

  6. ...xsin(x) - ∫ sin(x)dx ycos(x) = xsin(x) + cos(x) + C y = xtan(x) + Csec(x) + 1 answer//

    3 Answers · Science & Mathematics · 20/08/2011

  7. Just keep taking the derivatives until it isn't 0/0, you're allowed to do that. With the x^3 on the bottom you'll need to apply L'Hopital's rule 3 times to get this not to be zero (x^3 -> 3x^2 -> 6x -> 6.)

    1 Answers · Science & Mathematics · 22/04/2009

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