Posted by jackie on Tuesday, December 8, 2009 at 6:26pm.
f(x)=cot(3x)csc^2(3x) = cosx/sin^3x
Let sinx = u . Then du = cosx dx
The integral of f(x) becomes
Integral of du/u^3 = -(1/2) u^-2
= (-1/2)(1/sinx)^2
= (-1/2)csc^2x
first change it to
f(x) = cos 3x/(sin 3x)^3 like drwls had
now use the quotient rule to find f'(x)
f'(x) = [(sin 3x)^3(-3)(sin 3x) - 3(sin 3x)^2(cos 3x)(3)(cos 3x)]/(sin 3x)^6
= -3[(sin 3x)^2 + 3(cos 3x)^2]/(sin 3x)^4
I made at least one mistake. The first was using x instead of 3x.
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