find ∫_0 ^(π/4) ((x−1)/(2+cosx))dx .


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Question Number 40658 by math khazana by abdo last updated on 25/Jul/18

$f i n d \int_{0}^{\frac{π}{4}} \frac{x - 1}{2 + c o s x} d x .$
Commented by math khazana by abdo last updated on 26/Jul/18
$let I = ∫_0 ^(π/4) ((x−1)/(2+cosx))dx changement tan((x/2))=t give I = ∫_0 ^((√2)−1) ((2arctant−1)/(2+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = 2 ∫_0 ^((√2)−1) ((2arctant −1)/(2+2t^2 +1−t^2 ))dt = 2 ∫_0 ^((√2)−1) ((2arctan(t)−1)/(3+t^2 ))dt =4 ∫_0 ^((√2)−1) ((arctan(t))/(3+t^2 ))dt −2 ∫_0 ^((√2)−1) (dt/(3+t^2 )) but ∫_0 ^((√2)−1) (dt/(3+t^2 )) =_(t=(√3)u) ∫_0 ^(((√2)−1)/(√3)) (((√3)du)/(3(1+u^2 ))) =((√3)/3) [ arctan(u)]_0 ^(((√2)−1)/(√3)) =((√3)/3) arctan((((√2)−1)/(√3))). let f(x)= ∫_0 ^((√2)−1) ((arctan(xt))/(3 +t^2 )) dt we have f^′ (x)= ∫_0 ^((√2)−1) (t/((3+t^2 )(1+x^2 t^2 )))dt let decompose F(t) = (t/((3+t^2 )(1+x^2 t^2 ))) F(t) = ((at +b)/(t^2 +3)) + ((ct +d)/(x^2 t^2 +1)) F(−t)=−F(t)⇒ ((−at +b)/(t^2 +3)) +((−ct +d)/(x^2 t^2 +1)) =((−at−b)/(t^2 +3)) +((−ct−d)/(x^2 t^2 +1)) ⇒b=d=0 ⇒ F(t) = ((at)/(t^2 +3)) +((ct)/(x^2 t^2 +1)) lim_(t→+∞) tF(t) =0= a +(c/x^2 ) ⇒ax^2 +c =0 ⇒c=−ax^2 ⇒F(x) = ((at)/(t^2 +3)) −((ax^2 t)/(x^2 t^2 +1)) F(1) = (1/(4(1+x^2 ))) = (a/4) −((ax^2 )/(x^2 +1)) ⇒ 1=((4(1+x^2 )a)/4) −4(1+x^2 ) ((ax^2 )/(x^2 +1)) ⇒ 1=(1+x^2 )a −4ax^2 =(1+x^2 −4x^2 )a=(1−3x^2 )a ⇒a =(1/(1−3x^2 )) ⇒ F(x) = (1/(1−3x^2 )){ (t/(t^2 +3)) −((x^2 t)/(x^2 t^2 +1))}⇒ ∫_0 ^((√2)−1) F(t)dt =(1/(1−3x^2 )){ ∫_0 ^((√2)−1) ((tdt)/(t^2 +3)) −∫_0 ^((√2)−1) ((x^2 t)/(x^2 t^2 +1))dt} but ∫_0 ^((√2)−1) ((tdt)/(t^2 +3)) =(1/2)[ln(t^2 +3)]_0 ^((√2)−1) =(1/2){ln( ((√2)−1)^2 +3)−ln(3)} ∫_0 ^((√2)−1) ((x^2 t)/(x^2 t^2 +1))dt =[(1/2)ln(x^2 t^2 +1)]_(t=0) ^(t=(√2)−1) =(1/2){ ln(x^2 ((√2)−1)^2 +1) ∫_0 ^((√2)−1) F(t)dt = (1/(1−3x^2 )){ (1/2)ln(((√2)−1)^2 +3)−(1/2)ln(3)} −(1/2){ ln(((√2)−1)^2 x^2 +1)....be continued...$
$l e t I = \int_{0}^{\frac{π}{4}} \frac{x - 1}{2 + c o s x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t$ $g i v e I = \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n t - 1}{2 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n t - 1}{2 + 2 t^{2} + 1 - t^{2}} d t$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{2 a r c t a n (t) - 1}{3 + t^{2}} d t$ $= 4 \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (t)}{3 + t^{2}} d t - 2 \int_{0}^{\sqrt{2} - 1} \frac{d t}{3 + t^{2}} b u t$ $\int_{0}^{\sqrt{2} - 1} \frac{d t}{3 + t^{2}} =_{t = \sqrt{3} u} \int_{0}^{\frac{\sqrt{2} - 1}{\sqrt{3}}} \frac{\sqrt{3} d u}{3 (1 + u^{2})}$ $= \frac{\sqrt{3}}{3} {[a r c t a n (u)]}_{0}^{\frac{\sqrt{2} - 1}{\sqrt{3}}} = \frac{\sqrt{3}}{3} a r c t a n (\frac{\sqrt{2} - 1}{\sqrt{3}}) .$ $l e t f (x) = \int_{0}^{\sqrt{2} - 1} \frac{a r c t a n (x t)}{3 + t^{2}} d t$ $w e h a v e f^{^{'}} (x) = \int_{0}^{\sqrt{2} - 1} \frac{t}{(3 + t^{2}) (1 + x^{2} t^{2})} d t$ $l e t d e c o m p o s e F (t) = \frac{t}{(3 + t^{2}) (1 + x^{2} t^{2})}$ $F (t) = \frac{a t + b}{t^{2} + 3} + \frac{c t + d}{x^{2} t^{2} + 1}$ $F (- t) = - F (t) \Rightarrow \frac{- a t + b}{t^{2} + 3} + \frac{- c t + d}{x^{2} t^{2} + 1}$ $= \frac{- a t - b}{t^{2} + 3} + \frac{- c t - d}{x^{2} t^{2} + 1} \Rightarrow b = d = 0 \Rightarrow$ $F (t) = \frac{a t}{t^{2} + 3} + \frac{c t}{x^{2} t^{2} + 1}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + \frac{c}{x^{2}} \Rightarrow {a x}^{2} + c = 0 \Rightarrow c = - {a x}^{2}$ $\Rightarrow F (x) = \frac{a t}{t^{2} + 3} - \frac{{a x}^{2} t}{x^{2} t^{2} + 1}$ $F (1) = \frac{1}{4 (1 + x^{2})} = \frac{a}{4} - \frac{{a x}^{2}}{x^{2} + 1} \Rightarrow$ $1 = \frac{4 (1 + x^{2}) a}{4} - 4 (1 + x^{2}) \frac{{a x}^{2}}{x^{2} + 1} \Rightarrow$ $1 = (1 + x^{2}) a - 4 {a x}^{2} = (1 + x^{2} - 4 x^{2}) a = (1 - 3 x^{2}) a$ $\Rightarrow a = \frac{1}{1 - 3 x^{2}} \Rightarrow$ $F (x) = \frac{1}{1 - 3 x^{2}} {\frac{t}{t^{2} + 3} - \frac{x^{2} t}{x^{2} t^{2} + 1}} \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} F (t) d t = \frac{1}{1 - 3 x^{2}} {\int_{0}^{\sqrt{2} - 1} \frac{t d t}{t^{2} + 3} - \int_{0}^{\sqrt{2} - 1} \frac{x^{2} t}{x^{2} t^{2} + 1} d t}$ $b u t \int_{0}^{\sqrt{2} - 1} \frac{t d t}{t^{2} + 3} = \frac{1}{2} {[l n (t^{2} + 3)]}_{0}^{\sqrt{2} - 1}$ $= \frac{1}{2} {l n ({(\sqrt{2} - 1)}^{2} + 3) - l n (3)}$ $\int_{0}^{\sqrt{2} - 1} \frac{x^{2} t}{x^{2} t^{2} + 1} d t = {[\frac{1}{2} l n (x^{2} t^{2} + 1)]}_{t = 0}^{t = \sqrt{2} - 1}$ $= \frac{1}{2} {l n (x^{2} {(\sqrt{2} - 1)}^{2} + 1)$ $\int_{0}^{\sqrt{2} - 1} F (t) d t = \frac{1}{1 - 3 x^{2}} {\frac{1}{2} l n ({(\sqrt{2} - 1)}^{2} + 3) - \frac{1}{2} l n (3)}$ $- \frac{1}{2} {l n ({(\sqrt{2} - 1)}^{2} x^{2} + 1) . . . . b e c o n t i n u e d . . .$
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