Evaluate Integral: cot^{-1}(x^2 - x +1) from 0 to 1 | Integral Homework

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Homework Help Overview

The problem involves evaluating the integral \(\int^1_0 \cot^{-1}(x^2 - x + 1)\ dx\), which falls under the subject area of integral calculus. Participants are exploring various methods to approach the integral without arriving at a solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using integration by parts and substitution, noting that these methods have not yielded progress. There is mention of a specific formula involving \(2I\) and a transformation of the integrand into a form involving \(\tan^{-1}\). Some participants suggest comparing the problem to known identities or expansions.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and suggesting different approaches. Some hints have been provided, but there is no clear consensus or resolution yet. The exploration of various interpretations and methods continues.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the types of solutions they can pursue. There is an indication that some participants find the problem straightforward, while others are struggling with the methods.

utkarshakash
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Homework Statement


[itex]\displaystyle \int^1_0 cot^{-1}(x^2 - x +1)\ dx[/itex]

Homework Equations



The Attempt at a Solution


I used this formula

[itex]2I=\int^b_a f(x)+f(a+b-x)\ dx[/itex]

But using this method I arrived at the original question. OK, So I tried integrating by parts and it's still useless. Substitution doesn't work either.
 
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hi utkarshakash! :smile:

have you tried integrating by parts with u = x ?
 
utkarshakash said:

Homework Statement


[itex]\displaystyle \int^1_0 cot^{-1}(x^2 - x +1)\ dx[/itex]

Homework Equations



The Attempt at a Solution


I used this formula

[itex]2I=\int^b_a f(x)+f(a+b-x)\ dx[/itex]

But using this method I arrived at the original question. OK, So I tried integrating by parts and it's still useless. Substitution doesn't work either.

[tex]\cot^{-1}(x^2-x+1)=\tan^{-1}\left(\frac{1}{x^2-x+1}\right)=\tan^{-1}\left(\frac{1}{1-x(1-x)}\right)[/tex]

:smile:
 
In case Pranav-Arora's hint is still obscure, compare it with the expansion of tan(a+b).
 
utkarshakash said:

Homework Statement


[itex]\displaystyle \int^1_0 cot^{-1}(x^2 - x +1)\ dx[/itex]

Homework Equations



The Attempt at a Solution


I used this formula

[itex]2I=\int^b_a f(x)+f(a+b-x)\ dx[/itex]

But using this method I arrived at the original question. OK, So I tried integrating by parts and it's still useless. Substitution doesn't work either.

Firstly write it as,

tan-1(1/(x2-x+1) = tan-1{(x-(x-1))/(1+x(x-1))}

Don't you see an obvious relation of tan-1A - tan-1B formula from here ? You may proceed..

It was an easy question though. :smile:

Method II :

Let f(x) =cot−1(x2−x+1)

Write as,

f(x) =cot−1(x2−x+1)*1

Take cot−1(x2−x+1) as a first function as per ILATE, and 1 as second function. Then you can integrate by parts. :wink:
 

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