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By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

    Q8.    \int_0^\frac{\pi}{4}\log(1+\tan x)dx

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We have                       I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx

By using the identity

                                   \ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,       

                                    I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx

or                                I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx

or                                       I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx

or                                       I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx

or                                       I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I

or                                       2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0

or                                          I\ =\ \frac{\Pi }{8}\log2

Posted by

Devendra Khairwa

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