Question

Evaluate the integral:

(cos(2-3x) - tan(5-3x))dx
we know that
∫cos(u) du = sin(u)
∫tan(u) du = -log cos(u)

So, what we get here is
(-1/3)sin(2-3x) - (-1/3)(-log cos(5-3x))
= -1/3 (sin(2-3x) + log cos(5-3x)) + C
the correct answer I was given has a negative 1/3 before the log. Also, how did you get the 1/3 value?

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