Evaluate: ^ 1 _ 0 ^ -1 (1 - x + x^ 2 ) dx - Vidyadip

Question

Evaluate: $\int\limits^{1}_{0}\cot^{-1}(1 - x + x^{2}) dx$

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Answer

$\text{I} = \int\limits^{1}_{0} \cot^{-1}(1 - x + x^{2}) dx = \int\limits^{1}_{0}\tan^{-1}\bigg( \frac{1}{1 - x + x^{2}}\bigg)dx$
$=\int\limits^{1}_{0} \tan^{-1}\bigg(\frac{x + (1 - x)}{1 - x (1 - x)}\bigg)dx = \int\limits^{1}_{0}\tan^{-1}x\text{ }dx + \int\limits^{1}_{0}\tan^{-1}(1 - x) dx$
$ = 2\int\limits^{1}_{0}\tan^{-1} x\text{ }dx$
$= 2\Bigg[\bigg(\tan^{-1}x.x\bigg)^{1}_{0} - \int\limits^{1}_{0}\frac{x}{1 + x^{2}}dx\Bigg]$
$=2\bigg[x \tan^{-1} x - \frac{1} {2}\log|1 + x^{2}|\bigg]^{1}_{0}$
$= 2\bigg[\frac{\pi}{4}-\frac{1}{2} \log 2\bigg] \text{or} \frac{\pi}{2}-\log 2$

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