find ∫_(π/3) ^(π/2) (x/(cosx))dx


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Question Number 55996 by maxmathsup by imad last updated on 07/Mar/19

$f i n d \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{c o s x} d x$
Commented by maxmathsup by imad last updated on 09/Mar/19
$let I =∫_(π/3) ^(π/2) (x/(cosx))dx changement tan((x/2))=t give =∫_(1/(√3)) ^1 ((2arctan(t))/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) =4∫_(1/(√3)) ^1 ((arctant)/(1−t^2 )) dt but ∫_(1/(√3)) ^1 ((arctan(t))/(1−t^2 ))dt =(1/2)∫_(1/(√3)) ^1 arctan(t){(1/(1−t)) +(1/(1+t))}dt =(1/2) ∫_(1/(√3)) ^1 ((arctan(t))/(1−t))dt +(1/2) ∫_(1/(√3)) ^1 ((arctan(t))/(1+t))dt let f(x) =∫_(1/(√3)) ^1 ((arctan(xt))/(1−t)) dt ⇒ f^′ (x) = ∫_(1/(√3)) ^1 (t/((1+x^2 t^2 )(1−t)))dt =_(xt =u) ∫_(x/(√3)) ^x (u/(x(1+u^2 )(1−(u/x)))) (du/x) =(1/x) ∫_(x/(√3)) ^x (u/((u^2 +1)(x−u))) du let decompose F(u) =(u/((x−u)(u^2 +1))) F(u) =(a/(x−u)) +((bu +c)/(u^2 +1)) a =lim_(u→x) (x−u)F(u) =(x/(x^2 +1)) lim_(u→+∞) uF(u) =0 =−a +b ⇒b=a ⇒F(u) =(a/(x−u)) +((au +c)/(u^2 +1)) F(0) =0 =(a/x) +c ⇒c =−(a/x) =−(1/(x^2 +1)) ⇒ F(u) =(x/((x^2 +1)(x−u))) +(((x/(x^2 +1))u −(1/(x^2 +1)))/(u^2 +1)) =(x/(x^2 +1)){(1/(x−u)) +((u−x)/(u^2 +1))} ⇒ ∫ F(u)du =(x/(x^2 +1))ln∣x−u∣ +(x/(2(x^2 +1)))ln(u^2 +1) −(x^2 /(x^2 +1)) arctan(u) +c ⇒ ∫_(x/(√3)) ^x F(u)du =(x/(x^2 +1))[ln∣x−u∣+(1/2)ln(u^2 +1)−x arctanu]_(x/(√3)) ^x ....be continued....$
$l e t I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{x}{c o s x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 a r c t a n (t)}{\frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 4 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n t}{1 - t^{2}} d t b u t$ $\int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (t)}{1 - t^{2}} d t = \frac{1}{2} \int_{\frac{1}{\sqrt{3}}}^{1} a r c t a n (t) {\frac{1}{1 - t} + \frac{1}{1 + t}} d t$ $= \frac{1}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (t)}{1 - t} d t + \frac{1}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (t)}{1 + t} d t l e t f (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{a r c t a n (x t)}{1 - t} d t \Rightarrow$ $f^{^{'}} (x) = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t}{(1 + x^{2} t^{2}) (1 - t)} d t =_{x t = u} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{u}{x (1 + u^{2}) (1 - \frac{u}{x})} \frac{d u}{x}$ $= \frac{1}{x} \int_{\frac{x}{\sqrt{3}}}^{x} \frac{u}{(u^{2} + 1) (x - u)} d u l e t d e c o m p o s e F (u) = \frac{u}{(x - u) (u^{2} + 1)}$ $F (u) = \frac{a}{x - u} + \frac{b u + c}{u^{2} + 1}$ $a = {l i m}_{u \to x} (x - u) F (u) = \frac{x}{x^{2} + 1}$ ${l i m}_{u \to + \infty} u F (u) = 0 = - a + b \Rightarrow b = a \Rightarrow F (u) = \frac{a}{x - u} + \frac{a u + c}{u^{2} + 1}$ $F (0) = 0 = \frac{a}{x} + c \Rightarrow c = - \frac{a}{x} = - \frac{1}{x^{2} + 1} \Rightarrow$ $F (u) = \frac{x}{(x^{2} + 1) (x - u)} + \frac{\frac{x}{x^{2} + 1} u - \frac{1}{x^{2} + 1}}{u^{2} + 1}$ $= \frac{x}{x^{2} + 1} {\frac{1}{x - u} + \frac{u - x}{u^{2} + 1}} \Rightarrow \int F (u) d u = \frac{x}{x^{2} + 1} l n ∣ x - u ∣ + \frac{x}{2 (x^{2} + 1)} l n (u^{2} + 1)$ $- \frac{x^{2}}{x^{2} + 1} a r c t a n (u) + c \Rightarrow$ $\int_{\frac{x}{\sqrt{3}}}^{x} F (u) d u = \frac{x}{x^{2} + 1} {[l n ∣ x - u ∣ + \frac{1}{2} l n (u^{2} + 1) - x a r c t a n u]}_{\frac{x}{\sqrt{3}}}^{x}$ $. . . . b e c o n t i n u e d . . . .$
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