MHB Integration by Parts with Domain Warning

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The discussion focuses on the integration of the function $\int x \cot^2(x) \, dx$ using integration by parts. Participants agree on the choice of $u = x$ and $dv = \cot^2(x) \, dx$, leading to the calculation of $du$ and $v$. There is a request for clarification on the workings for $v$, which is derived from integrating $\cot^2(x)$ as $\int (\csc^2 x - 1) \, dx$. A participant mentions receiving a domain warning from their TI calculator during the process, indicating potential issues with the integration limits or function behavior. The discussion highlights the importance of careful integration steps and understanding domain restrictions.
karush
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$\int x\ \cot^2\left({x}\right) dx $
$u=x$ $dv=\cot^2\left({x}\right) dx $
$du=\frac{x^2}{2}$ $v=\frac{-\cos\left({x}\right)+x\sin\left({x}\right)}{\sin\left({x}\right)}$
 
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Hi karush,

Could you show your workings for $v$?
We have $dv=\cot^2\left({x}\right) dx$
$v=\int (\csc^2 x -1) \,dx$

Can you proceed?
 
I agree with you choice of $u$ and $dv$:

$$u=x\,\therefore\,du=dx$$ (You integrated rather than differentiated)

$$dv=\cot^2(x)\,dx=\left(\csc^2(x)-1\right)\,dx\,\therefore\,v=?$$
 
That what I got for $v$ with the TI ??
 
karush said:
That what I got for $v$ with the TI ??

I suspect your TI spat out:

$$v=-\frac{\cos\left({x}\right)+x\sin\left({x}\right)}{\sin\left({x}\right)}$$ :D
 
$uv-\int\ v\ du$

$-x(\cot\left({x}\right)-x)+\int \cot\left({x}\right)dx -\int x\ dx$

My TI returned a domain warning??
 
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