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December 18, 2014

December 18, 2014

Posted by **jackie** on Tuesday, December 8, 2009 at 6:26pm.

- calculus -
**drwls**, Tuesday, December 8, 2009 at 6:33pmf(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

- calculus -
**Reiny**, Tuesday, December 8, 2009 at 7:45pmfirst 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

- calculus -
**drwls**, Tuesday, December 8, 2009 at 9:47pmI made at least one mistake. The first was using x instead of 3x.

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