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

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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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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:
 
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: