calculate ∫_(−(π/4)) ^(π/4) (x^2 /(cos^2 x))dx


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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
$let I = ∫_(−(π/4)) ^(π/4) (x^2 /(cos^2 x))dx ⇒ I =∫_(−(π/4)) ^(π/4) (1+tan^2 x)x^2 dx changement tanx =t give I = ∫_(−1) ^1 (1+t^2 )(arctant)^2 (dt/(1+t^2 )) = 2∫_0 ^1 (arctant)^2 dt by parts I = 2{ [t (arctant)^2 ]_0 ^1 −∫_0 ^1 t ((2arctan(t))/(1+t^2 ))dt} =2{(π^2 /(16)) − ∫_0 ^1 ((2t arctan(t))/(1+t^2 ))} =(π^2 /8) −4 ∫_0 ^1 ((t arctan(t))/(1+t^2 )) dt let ϕ(x) =∫_0 ^1 ((t arctan(xt))/(1+t^2 )) dt ⇒ϕ^′ (x) = ∫_0 ^1 (t^2 /((1+x^2 t^2 )(1+t^2 )))dt ϕ^′ (x) =_(xt =u) ∫_0 ^x (u^2 /(x^2 (1+u^2 )(1+(u^2 /x^2 )))) (du/x) = ∫_0 ^x (u^2 /(x(1+u^2 )(x^2 +u^2 )))du =(1/x) ∫_0 ^1 (u^2 /((u^2 +1)( u^2 +x^2 )))du let decompose F(u) = (u^2 /((u^2 +1)(u^2 +x^2 ))) ⇒F(u) = ((au+b)/(u^2 +1)) +((cu +d)/(u^2 +x^2 )) F(−u) =F(u) ⇒((−au +b)/(u^2 +1)) +((−cu +d)/(u^2 +x^2 )) =F(u) ⇒a=c=0 ⇒ F(u) = (b/(u^2 +1)) +(d/(u^2 +x^2 )) lim_(u→+∞) u^2 F(u) =1 =b+d ⇒d=1−b ⇒ F(u) =(b/(u^2 +1)) +((1−b)/(u^2 +x^2 )) ⇒F(0) =0 =b +((1−b)/x^2 ) =(1/x^2 ) +(1−(1/x^2 ))b ⇒ −(1/x^2 ) =(((x^2 −1)b)/x^2 ) ⇒b =(1/(1−x^2 )) ⇒F(u) =(1/((1−x^2 )(u^2 +1))) +((1−(1/(1−x^2 )))/(u^2 +x^2 )) = (1/((1−x^2 )(u^2 +1))) −(x^2 /((1−x^2 )(u^2 +x^2 ))) ⇒F(u) =(1/(1−x^2 )){(1/(u^2 +1)) −(x^2 /(u^2 +x^2 ))} ϕ^′ (x) =(1/x)∫_0 ^x F(u)du =(1/(x(1−x^2 ))){ ∫_0 ^x (du/(1+u^2 )) −x^2 ∫_0 ^x (du/(u^2 +x^2 ))} =((arctanx)/(x(1−x^2 ))) −(x/(1−x^2 )) ∫_0 ^x (du/(u^2 +x^2 )) but ∫_0 ^x (du/(u^2 +x^2 )) du =_(u=xt) ∫_0 ^1 ((xdt)/(x^2 t^2 +x^2 )) =(1/x) ∫_0 ^1 (dt/(t^2 +1)) =(π/(4x)) ⇒ ϕ^′ (x) = ((arctan(x))/(x(1−x^2 ))) −(π/(4(1−x^2 ))) ( for x^2 ≠1) ⇒$
$l e t I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{2}}{{c o s}^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (1 + {t a n}^{2} x) x^{2} d x c h a n g e m e n t t a n x = t$ $g i v e I = \int_{- 1}^{1} (1 + t^{2}) {(a r c t a n t)}^{2} \frac{d t}{1 + t^{2}} = 2 \int_{0}^{1} {(a r c t a n t)}^{2} d t b y p a r t s$ $I = 2 {{[t {(a r c t a n t)}^{2}]}_{0}^{1} - \int_{0}^{1} t \frac{2 a r c t a n (t)}{1 + t^{2}} d t}$ $= 2 {\frac{π^{2}}{16} - \int_{0}^{1} \frac{2 t a r c t a n (t)}{1 + t^{2}}} = \frac{π^{2}}{8} - 4 \int_{0}^{1} \frac{t a r c t a n (t)}{1 + t^{2}} d t$ $l e t φ (x) = \int_{0}^{1} \frac{t a r c t a n (x t)}{1 + t^{2}} d t \Rightarrow φ^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{(1 + x^{2} t^{2}) (1 + t^{2})} d t$ $φ^{^{'}} (x) =_{x t = u} \int_{0}^{x} \frac{u^{2}}{x^{2} (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}$ $= \int_{0}^{x} \frac{u^{2}}{x (1 + u^{2}) (x^{2} + u^{2})} d u = \frac{1}{x} \int_{0}^{1} \frac{u^{2}}{(u^{2} + 1) (u^{2} + x^{2})} d u l e t d e c o m p o s e$ $F (u) = \frac{u^{2}}{(u^{2} + 1) (u^{2} + x^{2})} \Rightarrow F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + x^{2}}$ $F (- u) = F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + x^{2}} = F (u) \Rightarrow a = c = 0 \Rightarrow$ $F (u) = \frac{b}{u^{2} + 1} + \frac{d}{u^{2} + x^{2}}$ ${l i m}_{u \to + \infty} u^{2} F (u) = 1 = b + d \Rightarrow d = 1 - b \Rightarrow$ $F (u) = \frac{b}{u^{2} + 1} + \frac{1 - b}{u^{2} + x^{2}} \Rightarrow F (0) = 0 = b + \frac{1 - b}{x^{2}} = \frac{1}{x^{2}} + (1 - \frac{1}{x^{2}}) b \Rightarrow$ $- \frac{1}{x^{2}} = \frac{(x^{2} - 1) b}{x^{2}} \Rightarrow b = \frac{1}{1 - x^{2}} \Rightarrow F (u) = \frac{1}{(1 - x^{2}) (u^{2} + 1)} + \frac{1 - \frac{1}{1 - x^{2}}}{u^{2} + x^{2}}$ $= \frac{1}{(1 - x^{2}) (u^{2} + 1)} - \frac{x^{2}}{(1 - x^{2}) (u^{2} + x^{2})} \Rightarrow F (u) = \frac{1}{1 - x^{2}} {\frac{1}{u^{2} + 1} - \frac{x^{2}}{u^{2} + x^{2}}}$ $φ^{^{'}} (x) = \frac{1}{x} \int_{0}^{x} F (u) d u = \frac{1}{x (1 - x^{2})} {\int_{0}^{x} \frac{d u}{1 + u^{2}} - x^{2} \int_{0}^{x} \frac{d u}{u^{2} + x^{2}}}$ $= \frac{a r c t a n x}{x (1 - x^{2})} - \frac{x}{1 - x^{2}} \int_{0}^{x} \frac{d u}{u^{2} + x^{2}} b u t$ $\int_{0}^{x} \frac{d u}{u^{2} + x^{2}} d u =_{u = x t} \int_{0}^{1} \frac{x d t}{x^{2} t^{2} + x^{2}} = \frac{1}{x} \int_{0}^{1} \frac{d t}{t^{2} + 1} = \frac{π}{4 x} \Rightarrow$ $φ^{^{'}} (x) = \frac{a r c t a n (x)}{x (1 - x^{2})} - \frac{π}{4 (1 - x^{2})} (f o r x^{2} \neq 1) \Rightarrow$
Commented by maxmathsup by imad last updated on 02/Aug/18

$φ (x) = \int_{.}^{x} \frac{a r c t a n t}{t (1 - t^{2})} d t - \frac{π}{4} \int_{.}^{x} \frac{d t}{1 - t^{2}} + c b u t$ $\int \frac{d t}{1 - t^{2}} = \frac{1}{2} \int (\frac{1}{1 + t} + \frac{1}{1 - t}) d t = \frac{1}{2} l n ∣ \frac{1 + t}{1 - t} ∣ \Rightarrow$ $φ (x) = \int_{.}^{x} \frac{a r c t a n (t)}{t (1 - t^{2})} d t - \frac{π}{8} l n ∣ \frac{1 + x}{1 - x} ∣ + c . . . . b e c o n t i n u e d . . . .$
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