First slide

Evaluation of definite integrals

Question

$\int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x =$

Difficult

A

$\frac{π}{2}$
B

$\frac{π}{4}$
C

$\ln 2$
D

$\frac{π}{2} - \ln 2$

Solution

$\int_{0}^{1} \tan^{- 1} (1 - x + x^{2}) d x = \int_{0}^{1} \frac{π}{2} - \cot^{- 1} (1 - x + x^{2}) d x$
                                          $= \frac{π}{2} - \int_{0}^{1} \tan^{- 1} \frac{1}{1 - x + x^{2}} d x = \frac{π}{2} - \int_{0}^{1} \tan^{- 1} \frac{x + (1 - x)}{1 - x (1 - x)} d x$
                                          $= \frac{π}{2} - \int_{0}^{1} (\tan^{- 1} x + \tan^{- 1} (1 - x)) d x$
                              $= \frac{π}{2} - \int_{0}^{1} \tan^{- 1} x - \int_{0}^{1} \tan^{- 1} (1 - x) d x, x \to 1 - x = \frac{π}{2} - 2 \int_{0}^{1} \tan^{- 1} x d x$
                                          $= \frac{π}{2} - 2 [x {\tan^{- 1} x |}_{0}^{1} + {2 \int}_{0}^{1} \frac{x}{1 + x^{2}} d x$
                                          $= \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x = l {n (1 + x^{2}) |}_{0}^{1} = \ln 2$

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