find-the-value-of-0-pi-4-ln-1-tanx-dx-

Question Number 42391 by abdo.msup.com last updated on 24/Aug/18

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} l n (1 + t a n x) d x

Commented by maxmathsup by imad last updated on 25/Aug/18

l e t A = \int_{0}^{\frac{π}{4}} l n (1 + t a n x) d x c h a n g e m e n t x = \frac{π}{4} - t g i v e

A = \int_{0}^{\frac{π}{4}} l n (1 + t a n (\frac{π}{4} - t)) d t = \int_{0}^{\frac{π}{4}} l n (1 + \frac{1 - t a n t}{1 + t a n t}) d t

= \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n t}) d t = \frac{π}{4} l n (2) - \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t

= \frac{π}{4} l n (2) - A \Rightarrow 2 A = \frac{π}{4} l n (2) \Rightarrow A = \frac{π}{8} l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

I = \int_{0}^{\frac{Π}{4}} l n (1 + t a n x) d x

\int_{0}^{\frac{Π}{4}} l n {1 + t a n (\frac{Π}{4} - x)} d x

\int_{0}^{\frac{Π}{4}} l n {1 + \frac{1 - t a n x}{1 + t a n x}} d x

\int_{0}^{\frac{Π}{4}} l n (\frac{2}{1 + t a n x}) d x

= \int_{0}^{\frac{Π}{4}} l n 2 d x - \int_{0}^{\frac{Π}{4}} l n (1 + t a n x) d x

2 I = l n 2 \int_{0}^{\frac{Π}{4}} d x

I = l n 2 \times \frac{\frac{Π}{4}}{2} = l n 2 \times \frac{Π}{8}