Esther Tan	Posted on Tuesday, 22 April, 2003 - 10:45 am: Hi! could someone please help me with this integral? Hi! Could someone please help me with this integral?
Matthew Buckley	Posted on Tuesday, 22 April, 2003 - 11:25 am: Hi Esther, To integrate, think of it as ${\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cot }^{2}x{\cot }^{2}x\hspace{0.5em}\mathrm{dx},}$ and then change one of these by using the identity ${\displaystyle 1+{\cot }^{2}x=\backslash cosec{}^{2}x.}$ This will then give you two parts to integrate, and you should be able to do both of them. Anyway, hope thishelps -do post back if you are still stuck on the integration!
David Loeffler	Posted on Tuesday, 22 April, 2003 - 11:33 am: Oh no it's not! The answer is $\frac{\pi }{4}-\frac{2}{3}$. If you substitute $u=\cot x$, noting that ${\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}=-\backslash cosec{}^{2}x=-(1+u^2),}$ you get a rational function of $u$ that you can evaluate by partial fractions in the obvious way. The antiderivative (indefinite integral) is $x-\cot x+\frac{1}{3}{\cot }^{3}x.$ David
Esther Tan	Posted on Tuesday, 22 April, 2003 - 12:21 pm: David, i tried using your method but i end up with u2 /(u2 + 1) - u2 . Is this wrong, because i dont know how to split this into partial fractions! Matthew, when u say "This will then give you two parts to integrate", do you mean integrate cot2 x(cosec2 x -1) by parts or expand it to cot2 xcosec2 x - cot2 x and integrate that? either way, i dont know how to do both. Thanks for your help so far!
Matthew Buckley	Posted on Tuesday, 22 April, 2003 - 12:50 pm: Hi again Esther, Sorry for leaving you a bit stuck with the integration and for being a bit hazy. I shall explain further: You are correct so far. Expand to get cot2 xcosec2 x - cot2 x. These can both be integrated straight away as follows: Do you know what the differential of cot x is? If not, it is -cosec2 x. This will help. To integrate the cot2 x, again use the identity that cot2 x=cosec2 x - 1. As integration is the opposite to differentiation (roughly speaking,) then can you now see that cosec2 x integrates to -cot x? If not, post back. The 1 obviously just integrates to x. Now, the harder one - how to integrate cot2 xcosec2 x? Well, what happens if you differentiate cot2 x? Pull down the 2, leave the cot x as it is, reduce the power from 2 to 1, and multiply by the differential of cot x, which is -cosec2 x, so altogether, we get -2cotxcosec2 x, which nearly looks right, except we need a cot2 x, so try differentiating cot3 x, and see what you get. (You may have to fiddle a few constants and minus signs to get it exactly.) Then it is just matter of plugging in your limits. Does this help? If not, post back. P.S. Sorry if I don't tell you exactly all of what you need to do, but the best way to learn integration is just do loads of integrals by yourself, so if you do it, it will help you more, and thus you will get much better at integration! But do keep using us to keep nudging you further in the right direction if you are still unclear. P.P.S Formulae can also help when integrating things like tan2 x. You can use the identity tan2 x = sec2 x - 1, and sec2 x is nice, as it integrates nicely to tan x! Always try and rearrange what you are integrating into things that look like differentials of nice things. Hope that helps for any future problems you may encounter! P.P.P.S - Thank you Esther and David for clearing up the wrong limits! It looks a lot nicer now!
Philip Ellison	Posted on Tuesday, 22 April, 2003 - 12:58 pm: u2 /(u2 +1)=1-1/(u2 +1) This can then be solved using a further substitution.
David Loeffler	Posted on Tuesday, 22 April, 2003 - 01:02 pm: A slightly quicker method to do the partial fractions bit is to use difference of two squares: u4 = (u4 - 1) + 1 = (u2 + 1)(u2 - 1) + 1, so u4 /(u2 +1) = u2 - 1 + 1/(u2 + 1). David
Esther Tan	Posted on Tuesday, 22 April, 2003 - 01:07 pm: Ahhh..thank you so much Matthew. i forgot that cosec2 xcot2 x was in the form [f(x)]n f'(x) which can be simplified to 1/(n+1)[f(x)]n+1 + c and i also forgot that for top-heavy fractions u add a constant at the beginning! thanks Phillip! When i get 1-1/(u2 +1), i use u=tanx to get sec2 x right? thank you everyone for your kind help!
Esther Tan	Posted on Tuesday, 22 April, 2003 - 01:09 pm: I've never seen that before! very interesting David. thank u!