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

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I used this formula

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

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.

 

[tiny-tim]

hi utkarshakash!

have you tried integrating by parts with u = x ?

 

[Saitama]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I used this formula

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

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.

\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)

 

[haruspex]

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

 

[sankalpmittal]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I used this formula

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

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.

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.