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trig question

You can use cos(2x) = 1 - 2sin2x ==> sin2x = (1 - cos(2x)) /2 and the antiderative is x / 2 - 1/4 * (sin(2x)) + c 

If you take the derivative of  (Sinx)^3/(3cosx) +c, you get (1/3 cos(2 x)+2/3) tan^2(x). The thing is that when you use the quotient rule the denominator has its own derivative instead of 0. Say something like (6x + 5)7, the antiderivative is 1/48 (6x + 5)8, which the denominator is a constant and you can factor out and in the formula of quotient rule, d/dx (p(x)/q(x)) = (p'(x)q(x) - p(x)q'(x)) / (q(x))2, the p(x)q'(x) term would be 0 and this would simply becomes p'(x) / q(x), which is just a factoring of the constant denominator.

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