Integration by Parts with Domain Warning

In summary, the conversation discusses the use of integration by parts to solve the integral of $x\cot^2(x) dx$, with the choice of $u=x$ and $dv=\cot^2(x) dx$. The resulting expression for $v$ is derived to be $-\frac{\cos(x)+x\sin(x)}{\sin(x)}$, which is confirmed by the use of a TI calculator. There is also mention of a domain warning that may occur when using a TI calculator for this problem.
  • #1
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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  • #2
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?
 
  • #3
I agree with you choice of $u$ and $dv$:

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

\(\displaystyle dv=\cot^2(x)\,dx=\left(\csc^2(x)-1\right)\,dx\,\therefore\,v=?\)
 
  • #4
That what I got for $v$ with the TI ??
 
  • #5
karush said:
That what I got for $v$ with the TI ??

I suspect your TI spat out:

\(\displaystyle v=-\frac{\cos\left({x}\right)+x\sin\left({x}\right)}{\sin\left({x}\right)}\) :D
 
  • #6
$uv-\int\ v\ du$

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

My TI returned a domain warning??
 

FAQ: Integration by Parts with Domain Warning

What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It involves breaking down the original integral into two parts, and using the product rule of differentiation to simplify the integral.

When should I use integration by parts?

Integration by parts is most commonly used when the integral involves a product of two functions, and the integral cannot be directly evaluated using other methods such as substitution or trigonometric identities.

What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are the two functions involved in the integral. This formula is derived from the product rule of differentiation.

How do I choose which function to use as u and which one to use as dv?

In general, it is best to choose u as the function that will become simpler after differentiation, and dv as the function that will become easier to integrate. This may require some trial and error, and may also depend on the specific integral being evaluated.

Are there any special cases in integration by parts?

There are two special cases in integration by parts: when one of the functions is a polynomial, and when one of the functions is a logarithm. These cases have specific guidelines for choosing u and dv, and may require multiple iterations of the integration by parts formula.

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