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Question Number 41847 by maxmathsup by imad last updated on 13/Aug/18

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

1) f i n d a n o h e r e x p r e s s i o n o f f (x)

2) c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d t}{2 + t a n (t)} a n d A (θ) = \int_{0}^{\frac{π}{4}} \frac{d t}{s i n θ + t a n t}

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

Answered by maxmathsup by imad last updated on 14/Aug/18

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

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

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

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

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

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

x c = \frac{x^{2}}{1 + x^{2}} \Rightarrow c = \frac{x}{1 + x^{2}} (w e s u p p o s e x \neq 0) \Rightarrow

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

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

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

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

Commented by maxmathsup by imad last updated on 14/Aug/18

2) \int_{0}^{\frac{π}{4}} \frac{d t}{2 + t a n (t)} = f (2) = \frac{1}{5} {l n (3) - l n (2) - \frac{1}{2} l n (2) + \frac{π}{2}}

= \frac{1}{5} {l n (3) - \frac{3}{2} l n (2) + \frac{π}{2}} a l s o w e h a v e

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

Commented by maxmathsup by imad last updated on 14/Aug/18

3) w e h a v e f^{^{'}} (x) = - \int_{0}^{\frac{π}{4}} \frac{d t}{{(x + t a n t)}^{2}} \Rightarrow \int_{0}^{\frac{π}{4}} \frac{d t}{{(x + t a n t)}^{2}} = - f^{^{'}} (x) b u t

(1 + x^{2}) f (x) = l n ∣ 1 + x ∣ - l n ∣ x ∣ - \frac{l n (2)}{2} + \frac{π x}{4} b y d e r i v a t i o n

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

2 f (1) + 2 f^{^{'}} (1) = \frac{1}{2} - 1 + \frac{π}{4} = - \frac{1}{2} + \frac{π}{4} \Rightarrow f (1) + f^{^{'}} (1) = - \frac{1}{4} + \frac{π}{8} \Rightarrow

f^{^{'}} (1) = \frac{π}{8} - \frac{1}{4} - f (1) a n d

\int_{0}^{\frac{π}{4}} \frac{d t}{{(1 + t a n t)}^{2}} = - f^{^{'}} (1) = - \frac{π}{8} + \frac{1}{4} + f (1)

= - \frac{π}{8} + \frac{1}{4} + \frac{1}{2} {l n (2) - \frac{1}{2} l n (2) + \frac{π}{4}} = - \frac{π}{8} + \frac{1}{4} + \frac{l n (2)}{4} + \frac{π}{8}

= \frac{1}{4} + \frac{l n (2)}{4} .