calculate ∫_(π/2) ^(π/3) ((xdx)/(3+cosx))


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Question Number 68596 by Abdo msup. last updated on 14/Sep/19

$c a l c u l a t e \int_{\frac{π}{2}}^{\frac{π}{3}} \frac{x d x}{3 + c o s x}$
Commented by mathmax by abdo last updated on 15/Sep/19
$changeent tan((x/2))=t give I =∫_(π/2) ^(π/3) ((xdx)/(3+cosx)) =∫_1 ^(1/(√3)) ((2arctan(t))/(3+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_1 ^(1/(√3)) ((4arctan(t))/(3+3t^2 +1−t^2 ))dt =4 ∫_1 ^(1/(√3)) ((arctan(t))/(2t^2 +4))dt =2 ∫_1 ^(1/(√3)) ((arctan(t))/(t^2 +2))dt let f(α) =∫_1 ^(1/(√3)) ((arctan(αt))/(t^(2 ) +2))dt with α>0 ⇒ f^′ (α)=∫_1 ^(1/(√3)) (t/((α^2 t^2 +1)(t^2 +2)))dt let decompose F(t) =(t/((α^2 t^2 +1)(t^2 +2))) ⇒F(t)=((at+b)/(α^2 t^2 +1)) +((ct +d)/(t^2 +2)) F(−t)=−F(t) ⇒b=d=0 ⇒F(t) =((at)/(α^2 t^2 +1)) +((ct)/(t^2 +2)) lim_(t→+∞) tF(t) =0 =(a/α^2 ) +c ⇒c=−(a/α^2 ) ⇒ F(t) =((at)/(α^2 t^2 +1))−((at)/(α^2 (t^2 +2))) F(1) =(1/(3(α^2 +1))) =(a/(α^2 +1)) −(a/(3α^2 )) ⇒(1/3) =a−(a/(3α^2 ))(α^2 +1) ⇒ 1 =3a−((a(α^(2 ) +1))/α^2 ) =(((3α^2 −α^2 −1)a)/α^2 ) ⇒(2α^2 −1)a =α^2 ⇒ a =(α^2 /(2α^2 −1)) ⇒F(t) =((α^2 t)/((2α^2 −1)(α^2 t^2 +1))) −(t/((2α^2 −1)(t^2 +2))) ⇒ f^′ (α) =(1/(2(2α^2 −1))) ∫_1 ^(1/(√3)) ((2α^2 t)/(α^2 t^2 +1))dt−(1/(2(2α^2 −1))) ∫_1 ^(1/(√3)) ((2tdt)/(t^2 +2)) =(1/(2(2α^2 −1)))[ln(α^2 t^2 +1)]_1 ^(1/(√3)) −(1/(2(2α^2 −1))) [ln(t^(2 ) +2)]_1 ^(1/(√3)) =(1/(2(2α^2 −1))){ln((α^2 /3)+1)−ln(α^2 +1)}−(1/(2(2α^2 −1))){ln((1/3)+2)−ln(3)} ⇒f(α) =∫((ln(α^2 +3)−ln(α^2 +1)−ln(3))/(2(2α^2 −1)))dα −(ln((7/3))−ln(3))∫ (dα/(2(2α^2 −1))) +c....becontinued....$
$c h a n g e e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int_{\frac{π}{2}}^{\frac{π}{3}} \frac{x d x}{3 + c o s x} = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 a r c t a n (t)}{3 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{1}^{\frac{1}{\sqrt{3}}} \frac{4 a r c t a n (t)}{3 + 3 t^{2} + 1 - t^{2}} d t = 4 \int_{1}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n (t)}{2 t^{2} + 4} d t = 2 \int_{1}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n (t)}{t^{2} + 2} d t$ $l e t f (α) = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n (α t)}{t^{2} + 2} d t w i t h α > 0 \Rightarrow$ $f^{^{'}} (α) = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{t}{(α^{2} t^{2} + 1) (t^{2} + 2)} d t l e t d e c o m p o s e$ $F (t) = \frac{t}{(α^{2} t^{2} + 1) (t^{2} + 2)} \Rightarrow F (t) = \frac{a t + b}{α^{2} t^{2} + 1} + \frac{c t + d}{t^{2} + 2}$ $F (- t) = - F (t) \Rightarrow b = d = 0 \Rightarrow F (t) = \frac{a t}{α^{2} t^{2} + 1} + \frac{c t}{t^{2} + 2}$ ${l i m}_{t \to + \infty} t F (t) = 0 = \frac{a}{α^{2}} + c \Rightarrow c = - \frac{a}{α^{2}} \Rightarrow$ $F (t) = \frac{a t}{α^{2} t^{2} + 1} - \frac{a t}{α^{2} (t^{2} + 2)}$ $F (1) = \frac{1}{3 (α^{2} + 1)} = \frac{a}{α^{2} + 1} - \frac{a}{3 α^{2}} \Rightarrow \frac{1}{3} = a - \frac{a}{3 α^{2}} (α^{2} + 1) \Rightarrow$ $1 = 3 a - \frac{a (α^{2} + 1)}{α^{2}} = \frac{(3 α^{2} - α^{2} - 1) a}{α^{2}} \Rightarrow (2 α^{2} - 1) a = α^{2} \Rightarrow$ $a = \frac{α^{2}}{2 α^{2} - 1} \Rightarrow F (t) = \frac{α^{2} t}{(2 α^{2} - 1) (α^{2} t^{2} + 1)} - \frac{t}{(2 α^{2} - 1) (t^{2} + 2)} \Rightarrow$ $f^{^{'}} (α) = \frac{1}{2 (2 α^{2} - 1)} \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 α^{2} t}{α^{2} t^{2} + 1} d t - \frac{1}{2 (2 α^{2} - 1)} \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 t d t}{t^{2} + 2}$ $= \frac{1}{2 (2 α^{2} - 1)} {[l n (α^{2} t^{2} + 1)]}_{1}^{\frac{1}{\sqrt{3}}} - \frac{1}{2 (2 α^{2} - 1)} {[l n (t^{2} + 2)]}_{1}^{\frac{1}{\sqrt{3}}}$ $= \frac{1}{2 (2 α^{2} - 1)} {l n (\frac{α^{2}}{3} + 1) - l n (α^{2} + 1)} - \frac{1}{2 (2 α^{2} - 1)} {l n (\frac{1}{3} + 2) - l n (3)}$ $\Rightarrow f (α) = \int \frac{l n (α^{2} + 3) - l n (α^{2} + 1) - l n (3)}{2 (2 α^{2} - 1)} d α$ $- (l n (\frac{7}{3}) - l n (3)) \int \frac{d α}{2 (2 α^{2} - 1)} + c . . . . b e c o n t i n u e d . . . .$
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