find-0-pi-4-ln-1-tanx-dx-

Question Number 29451 by prof Abdo imad last updated on 08/Feb/18

f i n d \int_{0}^{\frac{π}{4}} l n (1 + t a n x) d x .

Commented by prof Abdo imad last updated on 28/Feb/18

l e t i n t r o d u c e t b e f u n c t i o n F (t) = \int_{0}^{\frac{π}{4}} l n (1 + t t a n x) d x

F^{^{'}} (t) = \int_{0}^{\frac{π}{4}} \frac{t a n x}{1 + t t a n x} d x (l e t t a k e t ⩾ o)

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

= \frac{π}{4 t} - \frac{1}{t} \int_{0}^{\frac{π}{4}} \frac{d x}{1 + t t a n x} b u t

\int_{0}^{\frac{π}{4}} \frac{d x}{1 + t t a n x} d x t h e c h . t a n x = u g i v e

\int_{0}^{\frac{π}{4}} \frac{d x}{1 + t t a n x} = \int_{0}^{1} \frac{1}{1 + t u} \frac{d u}{1 + u^{2}}

= \int_{0}^{1} \frac{d u}{(1 + u^{2}) (1 + t u)} d e c o m p o s i t i o n

F (u) = \frac{1}{(1 + u^{2}) (1 + t u)} = \frac{1}{t (u + \frac{1}{t}) (1 + u^{2})} \Rightarrow

t F (u) = \frac{a}{u + \frac{1}{t}} + \frac{b u + c}{1 + u^{2}} = \frac{t}{(1 + u^{2}) (1 + t u)}

a = {l i m}_{u \to - \frac{1}{t}} (u + \frac{1}{t}) t F (u) = \frac{1}{1 + \frac{1}{t^{2}}} = \frac{t^{2}}{1 + t^{2}}

{l i m}_{u \to \infty} u t F (u) = 0 = a + b \Rightarrow b = - a = - \frac{t^{2}}{1 + t^{2}}

t F (u) = \frac{t^{2}}{(1 + t^{2}) (u + \frac{1}{t})} + \frac{- \frac{t^{2}}{1 + t^{2}} u + c}{1 + u^{2}} b e c o n t i n u e d \dots .