First slide

Evaluation of definite integrals

Question

$\int_{0}^{\frac{π}{4}} x \tan x \tan (\frac{π}{4} - x) d x =$

Difficult

A

$\frac{π}{8} (\frac{π}{4} - \ln 2)$
B

$\frac{π}{8} (\frac{π}{4} + \ln 2)$
C

$\frac{π}{8} (\frac{π}{2} - \ln 2)$
D

$\frac{π}{4} (\frac{π}{2} + \ln 2)$

Solution

$I = \int_{0}^{\frac{π}{4}} x \tan x \tan (\frac{π}{4} - x) d x, x \to \frac{π}{4} - x$
$I = \int_{0}^{\frac{π}{4}} (\frac{π}{4} - x) \tan (\frac{π}{4} - x) \tan x d x$

$2 I = \frac{π}{4} \int_{0}^{\frac{π}{4}} \tan x \tan (\frac{π}{4} - x) d x$ ,

$I = \frac{π}{8} \int_{0}^{\frac{π}{4}} \tan x (\frac{1 - \tan x}{1 + \tan x}) d x, \tan x = t$

$I = \frac{π}{8} \int_{0}^{1} \frac{t (1 - t)}{(1 + t) (1 + t^{2})} d t$

$= \frac{π}{8} \int_{0}^{1} (\frac{1}{t^{2} + 1} - \frac{1}{t + 1}) d t$

$= \frac{π}{8} (\tan^{- 1} {t - \ln (t + 1) |}_{0}^{1}$

$= \frac{π}{8} (\frac{π}{4} - \ln 2)$

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