calculate-pi-4-pi-4-x-2-cos-2-x-dx-

Question Number 41049 by prof Abdo imad last updated on 01/Aug/18

c a l c u l a t e \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{2}}{{c o s}^{2} x} d x

Commented by maxmathsup by imad last updated on 02/Aug/18

l e t I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{2}}{{c o s}^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (1 + {t a n}^{2} x) x^{2} d x c h a n g e m e n t t a n x = t

g i v e I = \int_{- 1}^{1} (1 + t^{2}) {(a r c t a n t)}^{2} \frac{d t}{1 + t^{2}} = 2 \int_{0}^{1} {(a r c t a n t)}^{2} d t b y p a r t s

I = 2 {{[t {(a r c t a n t)}^{2}]}_{0}^{1} - \int_{0}^{1} t \frac{2 a r c t a n (t)}{1 + t^{2}} d t}

= 2 {\frac{π^{2}}{16} - \int_{0}^{1} \frac{2 t a r c t a n (t)}{1 + t^{2}}} = \frac{π^{2}}{8} - 4 \int_{0}^{1} \frac{t a r c t a n (t)}{1 + t^{2}} d t

l e t φ (x) = \int_{0}^{1} \frac{t a r c t a n (x t)}{1 + t^{2}} d t \Rightarrow φ^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{(1 + x^{2} t^{2}) (1 + t^{2})} d t

φ^{^{'}} (x) =_{x t = u} \int_{0}^{x} \frac{u^{2}}{x^{2} (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}

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

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

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

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

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

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

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

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

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

= \frac{a r c t a n x}{x (1 - x^{2})} - \frac{x}{1 - x^{2}} \int_{0}^{x} \frac{d u}{u^{2} + x^{2}} b u t

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

φ^{^{'}} (x) = \frac{a r c t a n (x)}{x (1 - x^{2})} - \frac{π}{4 (1 - x^{2})} (f o r x^{2} \neq 1) \Rightarrow

Commented by maxmathsup by imad last updated on 02/Aug/18

φ (x) = \int_{.}^{x} \frac{a r c t a n t}{t (1 - t^{2})} d t - \frac{π}{4} \int_{.}^{x} \frac{d t}{1 - t^{2}} + c b u t

\int \frac{d t}{1 - t^{2}} = \frac{1}{2} \int (\frac{1}{1 + t} + \frac{1}{1 - t}) d t = \frac{1}{2} l n ∣ \frac{1 + t}{1 - t} ∣ \Rightarrow

φ (x) = \int_{.}^{x} \frac{a r c t a n (t)}{t (1 - t^{2})} d t - \frac{π}{8} l n ∣ \frac{1 + x}{1 - x} ∣ + c \dots . b e c o n t i n u e d \dots .