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Integration of csc(x)^2 * cot(x)

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.








Cheers,

Patrick
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.
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pc0019
OP
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 ddxcotx=−cosec2x \frac{d}{dx} cot x = -cosec^2 x .

Now consider this: differentiate cot2x cot^2 x .

ddxcot2x=−2cosec2xcotx \frac{d}{dx} cot^2 x = -2cosec^2 x cot x

All were are now is a factor of −12 \frac{-1}{2} off the expression we had to integrate, therefore the value of the integral must be

−12cot2x \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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