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let-f-a-pi-4-pi-3-a-tan-2-x-dx-with-a-gt-0-1-find-a-explicit-form-of-f-a-2-find-also-g-a-pi-4-pi-3-dx-a-tan-2-x-3-find-the-values-of-pi-4-pi-3-2-tan-2-x-dx




Question Number 57665 by maxmathsup by imad last updated on 09/Apr/19
let f(a) =∫_(π/4) ^(π/3) (√(a+tan^2 x))dx   with a>0  1) find a explicit form of f(a)  2)  find also g(a) =∫_(π/4) ^(π/3)   (dx/( (√(a+tan^2 x))))  3) find the values of  ∫_(π/4) ^(π/3)  (√(2+tan^2 x))dx  and ∫_(π/4) ^(π/3)    (dx/( (√(3+tan^2 x))))
letf(a)=π4π3a+tan2xdxwitha>01)findaexplicitformoff(a)2)findalsog(a)=π4π3dxa+tan2x3)findthevaluesofπ4π32+tan2xdxandπ4π3dx3+tan2x
Commented by maxmathsup by imad last updated on 12/Apr/19
1)  changement tanx =(√a)u  give f(a) =∫_(1/( (√a))) ^((√3)/( (√a))) (√a)(√(1+u^2 ))  ((√a)/(1+au^2 )) du  =a ∫_(1/( (√a))) ^((√3)/( (√a)))    ((√(1+u^2 ))/(1+au^2 )) du  =_(u =sh(t))       a ∫_(argsh((1/( (√a))))) ^(argsh(((√3)/( (√a)))))    ((ch(t)cht dt)/(1+a sh^2 t))  =a ∫_(ln((1/( (√a))) +(√(1+(1/a))))) ^(ln(((√3)/( (√a))) +(√(1+(3/a)))))    (((1+ch(2t))/2)/(1+a ((ch(2t)−1)/2))) dt = a ∫_α ^β     ((1+ch(2t))/(2−a +ach(2t))) dt  =a ∫_α ^β    ((1+((e^(2t)  +e^(−2t) )/2))/(2−a +a ((e^(2t)  +e^(−2t) )/2))) dt = a ∫_α ^β       ((2 +e^(2t)  +e^(−2t) )/(4−2a +a e^(2t)   +a e^(−2t) )) dt  =_(e^(2t)  =u )          a∫_e^(2κ)  ^e^(2β)       ((2+u +u^(−1) )/(4−2a +au +au^(−1) )) (du/(2u))  =a ∫_(((1/( (√a))) +(√(1+(1/a))))^2 ) ^((((√3)/( (√a))) +(√(1+(3/a))))^2 )      ((2u +u^2  +1)/(2u^2 (4−2a) +2au^3  +2au)) du  =(a/2) ∫_((...)^2 ) ^((...)^2 )      ((u^2  +2u +1)/(u( au^2   +(4−2a)u +a))) du  let decompose  F(u) =((u^2  +2u +1)/(u(au^2  +(4−2a)u +a)))  roots of  p(u) =au^2  +(4−2a)u +a   Δ^′  =(2−a)^2 −a^2  =a^2  −4a +4 −a^2  =4(1−a)  case 1    0<a<1  ⇒ u_1 =((−2+a +2(√(1−a)))/a)  and u_2 =((−2+a −2(√(1−a)))/a)  F(u) =((u^2  +2u +1)/(au (u−u_1 )(u−u_2 ))) =(λ_0 /u) +(λ_1 /(u−u_1 )) +(λ_2 /(u−u_2 ))  λ_0 =lim_(u→0)  uF(u) =(1/a)  λ_1 =lim_(u→u_1 )   (u−u_1 )F(u) =((u_1 ^2  +2u_1  +1)/(au_1 4((√(1−a))/a))) =((u_1 ^2  +2u_1  +1)/(4u_1 (√(1−a))))  λ_2 =lim_(u→u_2  )  (u−u_2 )F(u) =((u_2 ^2  +2u_2  +1)/(au_2 (−4)((√(1−a))/a))) =−((u_2 ^2  +2u_2  +1)/(u_2 (√(1−a))))  rest to end calculus  ⇒ ∫ F(u)du =λ_0 ln∣u∣ +λ_1 ln∣u−u_1 ∣ +λ_2 ln∣u−u_2 ∣ +c ⇒
1)changementtanx=augivef(a)=1a3aa1+u2a1+au2du=a1a3a1+u21+au2du=u=sh(t)aargsh(1a)argsh(3a)ch(t)chtdt1+ash2t=aln(1a+1+1a)ln(3a+1+3a)1+ch(2t)21+ach(2t)12dt=aαβ1+ch(2t)2a+ach(2t)dt=aαβ1+e2t+e2t22a+ae2t+e2t2dt=aαβ2+e2t+e2t42a+ae2t+ae2tdt=e2t=uae2κe2β2+u+u142a+au+au1du2u=a(1a+1+1a)2(3a+1+3a)22u+u2+12u2(42a)+2au3+2audu=a2()2()2u2+2u+1u(au2+(42a)u+a)duletdecomposeF(u)=u2+2u+1u(au2+(42a)u+a)rootsofp(u)=au2+(42a)u+aΔ=(2a)2a2=a24a+4a2=4(1a)case10<a<1u1=2+a+21aaandu2=2+a21aaF(u)=u2+2u+1au(uu1)(uu2)=λ0u+λ1uu1+λ2uu2λ0=limu0uF(u)=1aλ1=limuu1(uu1)F(u)=u12+2u1+1au141aa=u12+2u1+14u11aλ2=limuu2(uu2)F(u)=u22+2u2+1au2(4)1aa=u22+2u2+1u21aresttoendcalculusF(u)du=λ0lnu+λ1lnuu1+λ2lnuu2+c
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 .
f(a)=a2[λ0lnu+λ1lnuu1+λ2lnuu2](1+1+aa)2(3+3+aa)2=aλ0ln(3+3+aa)+aλ12ln(3+3+aa)2u1+aλ22ln(3+3+aa)2u2aλ0ln(1+1+aa)aλ12ln(1+1+aa)2u1aλ22ln(1+1+aa)2u2.λianduiareknown.
Commented by maxmathsup by imad last updated on 12/Apr/19
2) rest to find f(a) if a>1  ....  we have f^′ (a) =∫_(π/4) ^(π/3)   (dx/(2(√(a+tan^2 x)))) =(1/2)g(a) ⇒g(a) =2f^′ (a)....
2)resttofindf(a)ifa>1.wehavef(a)=π4π3dx2a+tan2x=12g(a)g(a)=2f(a).
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 } .
3)letbeginfromthebeginingwehaveA=π4π32+tan2xdxchangementtanx=2ugiveA=123221+u221+2u2du=212321+u21+2u2du=u=sh(t)2argsh(12)argsh(32)ch(t)ch(t)dt1+2sh2(t)=2ln(12+1+12)ln(32+1+32)1+ch(2t)21+2ch(2t)12dt=2ln(1+32)ln(3+52)1+ch(2t)2+2ch(2t)2dt=2αβ1+e2t+e2t2e2t+e2t2dt=2αβ2+e2t+e2te2t+e2tdt=e2t=u2e2αe2β2+u+u1u+u1du2u=(1+32)2(3+52)22u+u2+1u(u2+1)duletdecomposeF(u)=u2+2u+1u(u2+1)F(u)=au+bu+cu2+1a=limu0uF(u)=1limu+uF(u)=1=a+bb=0F(u)=1u+cu2+1F(1)=2=1+c2c2=1c=2F(u)=1u+21+u2F(u)du=lnu+2arctan(u)+c(1+32)2(3+52)2u2+2u+1u(u2+1)du=[lnu+2arctan(u)](1+32)2(3+52)2A=2ln(3+52)+2arctan{(3+52)2}2ln(1+32)2arctan{(1+32)2}.

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