Question Number 57665 by maxmathsup by imad last updated on 09/Apr/19

Commented by maxmathsup by imad last updated on 12/Apr/19

Commented by maxmathsup by imad last updated on 12/Apr/19
![⇒ f(a) =(a/2)[ λ_0 ln∣u∣ +λ_1 ln∣u−u_1 ∣ +λ_2 ln∣u−u_2 ∣ ]_((((1+(√(1+a)))/( (√a))))^2 ) ^(((((√3) +(√(3+a)))/( (√a))))^2 ) =aλ_0 ln((((√3) +(√(3+a)))/( (√a)))) +((aλ_1 )/2)ln∣((((√3)+(√(3+a)))/( (√a))))^2 −u_1 ∣ +((aλ_2 )/2)ln∣ ((((√3)+(√(3+a)))/( (√a))))^2 −u_2 ∣ −aλ_0 ln(((1+(√(1+a)))/( (√a)))) −((aλ_1 )/2)ln∣ (((1+(√(1+a)))/( (√a))))^2 −u_1 ∣ −((aλ_2 )/2)ln∣(((1+(√(1+a)))/( (√a))))^2 −u_2 ∣ . λ_i and u_i are known .](https://www.tinkutara.com/question/Q57793.png)
Commented by maxmathsup by imad last updated on 12/Apr/19

Commented by maxmathsup by imad last updated on 12/Apr/19
![3) let begin from the begining we have A =∫_(π/4) ^(π/3) (√(2+tan^2 x))dx changement tanx =(√2)u give A =∫_(1/( (√2))) ^((√3)/( (√2))) (√2)(√(1+u^2 )) ((√2)/(1+2u^2 )) du =2 ∫_(1/( (√2))) ^((√3)/( (√2))) ((√(1+u^2 ))/(1+2u^2 )) du =_(u =sh(t)) 2 ∫_(argsh((1/( (√2))))) ^(argsh(((√3)/( (√2))))) ((ch(t) ch(t)dt)/(1+2sh^2 (t))) =2 ∫_(ln((1/( (√2))) +(√(1+(1/2))))) ^(ln(((√3)/( (√2))) +(√(1+(3/2))))) (((1+ch(2t))/2)/(1+2 ((ch(2t)−1)/2))) dt = 2 ∫_(ln(((1+(√3))/( (√2))))) ^(ln((((√(3 )) +(√5))/( (√2))))) ((1+ch(2t))/(2+2ch(2t)−2))dt =2 ∫_α ^β ((1+((e^(2t) +e^(−2t) )/2))/((e^(2t) +e^(−2t) )/2)) dt =2 ∫_α ^β ((2 +e^(2t) +e^(−2t) )/(e^(2t) +e^(−2t) )) dt =_(e^(2t) =u) 2 ∫_e^(2α) ^e^(2β) ((2 +u +u^(−1) )/(u +u^(−1) )) (du/(2u)) = ∫_((((1+(√3))/( (√2))))^2 ) ^(((((√3) +(√5))/( (√2))))^2 ) ((2u+u^2 +1)/(u(u^2 +1))) du let decompose F(u) =((u^2 +2u +1)/(u(u^2 +1))) F(u) = (a/u) +((bu +c)/(u^2 +1)) a =lim_(u→0) uF(u) =1 lim_(u→+∞) u F(u) =1 =a+b ⇒b =0 ⇒F(u) =(1/u) +(c/(u^2 +1)) F(1) =2 = 1+(c/2) ⇒(c/2) =1 ⇒c =2 ⇒ F(u) =(1/u) +(2/(1+u^2 )) ⇒ ∫ F(u)du =ln∣u∣ +2arctan(u)+c ⇒ ∫_((((1+(√3))/( (√2))))^2 ) ^(((((√3) +(√5))/( (√2))))^2 ) ((u^2 +2u +1)/(u(u^2 +1))) du =[ln∣u∣ +2arctan(u)]_((((1+(√3))/( (√2))))^2 ) ^(((((√3) +(√5))/( (√2))))^2 ) ⇒ A=2ln((((√3) +(√5))/( (√2)))) +2arctan{((((√3) +(√5))/( (√2))))^2 }−2ln(((1+(√3))/( (√2))))−2 arctan{(((1+(√3))/( (√2))))^2 } .](https://www.tinkutara.com/question/Q57795.png)