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    • Thread Starter

    I've tried integrating csc(x)^2 * cot(x) w.r.t x by parts and substitution and get -0.5 * cot(x) both times, but the answer according to my question sheet and Maple is -0.5 * csc(x).

    Can someone please tell me what I'm doing epically wrong here. I'm guessing it's something really obvious and embarrassing!

    My working for both is below.



    You are correct, that  \displaystyle\int \csc^2 x \cot x \ \mathrm{d}x = -\frac{1}{2} \cot^2 x \ (+ \ C) . Apparently both your question sheet and Maple are wrong...

    edit: as IrrationalNumber said, however if your question sheet and Maple had:  \displaystyle\int \csc^2 x\cot x \ \mathrm{d}x = -\frac{1}{2} \csc^2 x \ (+ \ C) , then this is correct, as  \cot^2 x = \csc^2 x - 1 , and as 1 is just constant, it is accounted for in the '+ C', merely the C in the two forms are different.

    I haven't checked your working through, but if you mean the answer according to maple is -1/2 cosec^2(x) then they're the same thing. Remember the arbitrary constant and try to show that -1/2 cosec^2(x) = -1/2 cot^2(x) + C for some C.

    Another way of completing the question : write out what the integrand is in terms of cosines and sines and make a substitution.
    • Thread Starter

    Thanks both if you, I've never come across a situation like this before but it makes complete sense :]

    This question can be done in one line using a method called 'presence of the derivative'. Its little known but is very powerful. Basically you look in your expression for a term that, when differentiated, gives the other. In this case  \frac{d}{dx} cot x = -cosec^2 x .

    Now consider this: differentiate  cot^2 x .

     \frac{d}{dx} cot^2 x = -2cosec^2 x cot x

    All were are now is a factor of  \frac{-1}{2} off the expression we had to integrate, therefore the value of the integral must be

     \frac{-1}{2}cot^2 x

    Skips out a lot of algebra and saves on any mistakes that could be made there!
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