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Question Number 48717 by Abdo msup. last updated on 27/Nov/18

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

1) f i n d f (x) a t a s i m p l e f o r m

2) c a l c u l a t e \int_{0}^{\frac{π}{4}} l n (1 + 2 t a n (t)) d t

Commented by maxmathsup by imad last updated on 01/Dec/18

1) w e h s v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{t a n t}{1 + x t a n t} d t = \frac{1}{x} \int_{0}^{\frac{π}{4}} \frac{1 + x t a n t - 1}{1 + x t a n (t)} d t (x \neq o)

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

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

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

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

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

F (u) = \frac{x^{2}}{(x^{2} + 1) (x u + 1)} + \frac{- \frac{x}{1 + x^{2}} u + c}{u^{2} + 1}

F (0) = 1 = \frac{x^{2}}{x^{2} + 1} + c \Rightarrow c = 1 - \frac{x^{2}}{x^{2} + 1} = \frac{1}{x^{2} + 1} \Rightarrow

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

\int_{0}^{1} \frac{d u}{(1 + u^{2}) (1 + x u)} = \frac{x^{2}}{(x^{2} + 1)} \int_{0}^{1} \frac{d u}{x u + 1} - \frac{x}{2 (x^{2} + 1)} \int_{0}^{1} \frac{2 u}{u^{2} + 1} d u + \frac{1}{1 + x^{2}} \int_{0}^{1} \frac{d u}{1 + u^{2}}

= \frac{x}{x^{2} + 1} {[l n ∣ x u + 1 ∣]}_{u = 0}^{1} - \frac{x}{2 (x^{2} + 1)} {[l n (u^{2} + 1)]}_{u = 0}^{1} + \frac{1}{1 + x^{2}} {[a r c t a n u]}_{u = 0}^{1}

= \frac{x}{x^{2} + 1} l n ∣ x + 1 ∣ - \frac{x l n (2)}{2 (x^{2} + 1)} + \frac{π}{4 (1 + x^{2})} \Rightarrow f^{^{'}} (x) = \frac{x}{x^{2} + 1} l n ∣ x + 1 ∣ - \frac{x l n (2)}{2 (x^{2} + 1)}

+ \frac{π}{4 (1 + x^{2})} \Rightarrow f (x) = \int \frac{x l n ∣ x + 1 ∣}{x^{2} + 1} d x - \frac{l n (2)}{4} l n (x^{2} + 1) + \frac{π}{4} a r c t a n x + c

i f w e s u p p o s e x > - 1 b y p a r t s u^{^{'}} = \frac{x}{x^{2} + 1} a n d v = l n (x + 1) w e g e t

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