integration from 0 to 1 cot inverse (1-x+x squrare) dx - Maths - Integrals

Question

integration from 0 to 1
cot inverse (1-x+x squrare) dx

Santosh Kumar · Accepted Answer

Let

$I = \int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) d x = \int_{0}^{1} \tan^{- 1} (\frac{1}{1 - x + x^{2}}) d x = \int_{0}^{1} \tan^{- 1} {\frac{1}{1 - x (1 - x)}} d x = \int_{0}^{1} \tan^{- 1} {\frac{x + (1 - x)}{1 - x (1 - x)}} d x = \int_{0}^{1} {{\tan^{- 1} x + \tan^{- 1} (1 - x)} d x}^{} = \int_{0}^{1} {(\tan^{- 1} x) d x +}^{} \int_{0}^{1} \tan^{- 1} {(1 - x) d x}^{} = \int_{0}^{1} {(\tan^{- 1} x) d x + = \int_{0}^{1} \tan^{- 1} {{1 - (1 - x)} d x}^{}}^{} {∵ \int_{0}^{a} f (x) d x = = \int_{0}^{a} f (a - x) d x}$

$= \int_{0}^{1} \tan^{- 1} x \cdot d x + \int_{0}^{1} \tan^{- 1} x \cdot d x = 2 \int_{0}^{1} \tan^{- 1} x \cdot d x = 2 \int_{0}^{1} \tan^{- 1} x \cdot 1 \cdot d x = 2 [x \tan^{- 1} x]_{0}^{1} - 2 \int_{0}^{1} (\frac{x}{1 + x^{2}}) d x = 2 [x \tan^{- 1} x]_{0}^{1} - \int_{0}^{1} {(\frac{2 x}{1 + x^{2}})}^{} d x = 2 (\frac{π}{4} - 0) - [\log (1 + x^{2})]_{0}^{1} = \frac{π}{2} - (\log 2 - \log 1) = \frac{π}{2} - \log 2$

integration from 0 to 1..cot inverse (1-x+x squrare) dx

integration from 0 to 1..
cot inverse (1-x+x squrare) dx