How Do You Solve the Integral of t * csc^2(t) dt?

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Homework Help Overview

The problem involves finding the integral of the function t * csc^2(t) with respect to t, which falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts integration by parts but expresses confusion regarding the antiderivative of -cot(t). Some participants suggest using integration by parts as well and hint at expressing cotangent in terms of other trigonometric functions for simplification. Others note the presence of singularities in the integral and mention symmetry properties that could affect the evaluation.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to the integral. Some hints have been provided regarding trigonometric identities and substitution methods, but no consensus or complete solution has emerged yet.

Contextual Notes

Participants have noted the presence of singularities in the integral at specific points, which may influence the evaluation. There is also mention of potential constraints related to the limits of integration, although these have not been fully clarified.

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Homework Statement


integral (t * csc^2 (t) ) dt


Homework Equations





The Attempt at a Solution



t * -cot (t) - integral( 1 * -cot (t)) u= t dv= csc^2(t)
du= 1 v= - cot (t)
-t * cot(t) - ? I don't understand how to find the antiderivative of -cot(t)
 
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Integration by parts is the way I'd also go about this.

Hint: In terms of other trig functions, what is cotangent equal to? You should end up with something that is solvable by substitution.
 
The integral contains singularities whenever sin(x)=0, or x=n PI

If it's an integral over -T, to T then the integral is zero (by symmetry)
 
[tex]\cot x = \frac{\cos x}{\sin x}[/tex]

[tex]\int \cot x dx = \int \frac{\cos x}{\sin x} dx[/tex].

let u= sin x, then du = cos x dx

[tex]\int \frac{1}{u} du = \ln u + C = \ln (\sin x) + C[/tex]
 

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