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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...2012-11-18
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    You can differentiate -cot x+C2012-11-18
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    I agree with @J.M.: this belongs to any table of elementary primitives that you should learn by heart.2012-11-18
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    @J.M. I know that... I was interested in a proof that did not require previously knowing the derivative of cot x...2012-11-18
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    Then you should mention that little detail in your question. :)2012-11-18
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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)2012-11-18
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    @J.M. Edited. =P2012-11-18
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    Stop calling it a table of integrals and call it what it *really* is: a table of antiderivatives - intended to be used with the fundamental theorem of calculus. Then it'll become apparent (hopefully) that you verify the formulas in the table by differentiating the RHS and checking that you indeed get the LHS.2012-11-18
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    @kahen I'm not American... And I know all that... =P (I've edited my question to clarify what I am interested in here.)2012-11-18
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    @kahen But I don't WANT to do that... It's not that I don't believe it's true. I want to know if it's possible to integrate it some other way. (Or perhaps that why they include it in the table of integrals/antiderivatives... It's not easily provable without differentiation...)2012-11-18
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    Alright, if a (Weierstrass-like) substitution is good enough for you: let $x=\arctan\,u$...2012-11-18
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    This is not working as nicely as I hoped... I filled a page with triangles and multiple substitutions and it keeps getting more and more horrible. I can see why you suggested differentiating it! (Maybe I am just dumb...)2012-11-18

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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.)2012-11-19
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    I'm also learning calc and making the connection with powers of sin was a really good idea.2017-02-10
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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 $t=\tan\,x$...2012-11-18
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    I would upvote this if I had enough reputation. Thanks.2012-11-19
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    @J.M. Thanks, I've added this in.2012-11-19
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    Okay, upvoted. :) @davie: you should now have sufficient rep for voting...2012-11-19
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Dividing both numerator and denominator by $cos^2(x)$:

$\displaystyle\int{\csc^2(x)dx} = \int \dfrac{1}{\sin^2}dx = \int \dfrac{\sec^2(x)}{\tan^2(x)}dx$

Using substitution $u = tan(x)$, $du = sec^2(x)dx$:

$\displaystyle\int{\csc^2(x)dx} = \int \dfrac{1}{u^2}du = -\dfrac{1}{u} = -\dfrac{1}{\tan(x)}=-\cot(x)$

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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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Question: $\displaystyle\int{\csc^2(x)dx} = $

Using substitution: $u = \dfrac{1}{\sin^2(x)}$

$\dfrac{du}{dx} = \dfrac{-2\cos(x)}{\sin^3(x)} \quad \longrightarrow du = \left(\dfrac{-2\cos(x)}{\sin(x)}\right)\times \left(\dfrac{1}{\sin^2(x)}\right)dx$

$\therefore -\dfrac{1}{2}\tan(x)du = \dfrac{1}{\sin^2(x)} dx \qquad (1)$

We still must find what is $\tan(x)$ in terms of $u$, hence:

$u = \dfrac{\sin^2(x) + \cos^2(x)}{\sin^2(x)} = 1+ \cot^2(x)$

$u-1 = \cot^2(x)$

$\sqrt{u-1} = \cot(x)$

$\dfrac{1}{\sqrt{u-1}} = \tan(x)$

Therefore substitute in $(1)$ and our integral becomes:

$-\dfrac{1}{2}\displaystyle\int{(u-1)^{-\frac{1}{2}}}du = -\dfrac{1}{2}\dfrac{\sqrt{(u-1)}}{\dfrac{1}{2}} + C = -\sqrt{\left(\dfrac{1}{\sin^2(x)} - 1)\right)} + C $

$= -\sqrt{\left(\dfrac{1-\sin^2(x)}{\sin^2(x)}\right)} + C = -\cot(x) + C$

Hope this makes sense,