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According to my sheet of standard integrals,

$\int \csc^2x \, dx = -\cot x + C$.

I am interested in a proof for the integral of $\operatorname{cosec}^2x$ that does not require differentiating $\cot x$. (I already know how to prove it using differentiation, but am interested in how one would calculate it without differentiation.)

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If you know how to differentiate the cotangent, then you know how to integrate the result of differentiating the cotangent... –  J. M. Nov 18 '12 at 13:28
    
You can differentiate -cot x+C –  Amr Nov 18 '12 at 13:28
    
I agree with @J.M.: this belongs to any table of elementary primitives that you should learn by heart. –  Siminore Nov 18 '12 at 13:29
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Taking the derivative of $\cot$ is trivial though. It's by definition $1\over\tan$, so you just use the quotient rule. And then use that $\csc = {1\over\sin}$ (again by definition). (You silly Americans and your need to have separate names even for the reciprocals of the trigonometric functions :D) –  kahen Nov 18 '12 at 13:33
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Alright, if a (Weierstrass-like) substitution is good enough for you: let $x=\arctan\,u$... –  J. M. Nov 18 '12 at 13:43

3 Answers 3

up vote 3 down vote accepted

If you want to be that perverse. I learned a way to integrate a power of sine, so why not apply it to the $-2$ power? Keep one factor of sine, convert all others to cosine, substitute $u=\cos x$. If we do that here, we get $$ \int\frac{dx}{\sin^2 x} = \int \frac{\sin x dx}{\sin^3 x} =\int\frac{\sin x dx}{(1-\cos^2 x)^{3/2}} =\int\frac{-du}{(1-u^2)^{3/2}} . $$ Then we can evaluate this integral (somehow, maybe even a trig substitution) to get $$ \int\frac{-du}{(1-u^2)^{3/2}} = \frac{-u}{\sqrt{1-u^2}} + C = \frac{- \cos x}{\sin x} + C $$

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That was genius. I knew how to integrate powers of sine, but I didn't think to apply it here. Thanks. (When I get enough reputation I will upvote this.) –  daviewales Nov 19 '12 at 2:38

Alright, we could attempt a Weierstrass substitution if that's the sort of thing you want. Let $t=\tan(\frac{x}{2})$. Thus $\csc(x)=\frac{1+t^{2}}{2t}$ and $dx=\frac{2dt}{1+t^{2}}$. Therefore we have the following:
$$\int \csc^{2}(x)dx=\int \frac{1+t^{2}}{2t}\cdot\frac{1+t^{2}}{2t}\cdot\frac{2dt}{1+t^{2}}=\int\frac{1+t^{2}}{2t^{2}}dt=\frac{1}{2}\int (t^{-2}+1) dt$$ $$=\frac{1}{2} \left[ \frac{-1}{t}+t\right]+C=\frac{t^{2}-1}{2t}+C=\frac{\tan^{2}(x/2)-1}{2\tan(x/2)}+C$$
Which leads to the given result by application of the double angle formula.

Letting $u=\tan(x)$ works too. We get $\csc^{2}(x)=1+\frac{1}{u^{2}}=\frac{1+u^{2}}{u^{2}}$, and $\frac{du}{1+u^{2}}=dx$. Therefore the integral is $$\int \csc^{2}(x)=\int \frac{1+u^{-2}}{1+u^{2}}du=\int \frac{1}{u^{2}}du=\frac{-1}{\tan(x)}+C=-\cot(x)+C$$

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Things are quite simpler if you let $t=\tan\,x$... –  J. M. Nov 18 '12 at 23:49
    
I would upvote this if I had enough reputation. Thanks. –  daviewales Nov 19 '12 at 2:25
    
@J.M. Thanks, I've added this in. –  Daniel Littlewood Nov 19 '12 at 18:18
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Okay, upvoted. :) @davie: you should now have sufficient rep for voting... –  J. M. Nov 19 '12 at 22:47
    
Thanks @J.M. =D –  daviewales Nov 20 '12 at 10:45

Use the trig identity for $\text{csc}^2x$ and write the expression in terms of $\text{sin }x$ and $\text{cos }x$: $$ \int{\text{csc}^2x\text{ }dx}=\int{(\text{cot}^2x+1)\text{ }dx}=\int{\text{cot}^2x\text{ }dx}+\int{dx}=\int{\frac{\text{cos}^2x}{\text{sin}^2x}\text{ }dx}+\int{dx} $$ Prepare for integration by parts: $$ u=\text{cos }x~~~~~~~~~~du=-\text{ sin }x\text{ }dx~~~~~~~~~~dv=\frac{\text{cos }x}{\text{sin}^2x}\text{ }dx~~~~~~~~~~v=-\frac{1}{\text{sin }x} $$ Integrate by parts and simplify: $$ \int{\frac{\text{cos}^2x}{\text{sin}^2x}\text{ }dx}+\int{dx}=-\frac{\text{cos }x}{\text{sin }x}-\int{\frac{-\text{ sin }x}{-\text{ sin }x}\text{ }dx}+\int{dx}=-\text{ cot }x+C $$

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