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Question Number 54936 by maxmathsup by imad last updated on 14/Feb/19

let f(θ) = ∫_0 ^1  (√(x^2  +2(cosθ)x +1))dx   with θ ∈ R .  1) calculate f(θ)  2) find the value of g(θ)=∫_0 ^1    ((xsinθ)/(√(x^2  +2cosθ x +1)))dx

letf(θ)=01x2+2(cosθ)x+1dxwithθR.1)calculatef(θ)2)findthevalueofg(θ)=01xsinθx2+2cosθx+1dx

Commented by Abdo msup. last updated on 16/Feb/19

1) we have f(θ)=∫_0 ^1 (√(x^(2 ) +2cosθ x +1))dx  =∫_0 ^1 (√((x+cosθ)^2  +sin^2 θ))dθ  chang.x+cosθ=sinθ t  give f(θ)=∫_(coatanθ) ^(cotan((θ/2)))  ∣sinθ∣sinθ (√(1+t^2 ))dt  f(θ)=∣sinθ∣sinθ  ∫_(cotan(θ)) ^(cotan((θ/2))) (√(1+t^2 ))dt  let find ∫  (√(1+t^2 ))dt   changement t=sh(u)give  ∫  (√(1+t^2 ))dt =∫ ch(u)ch(u)du  =∫ ch^2 (u)du =∫  ((1+ch(2u))/2)du  =(u/2) +(1/4)sh(2u) =(u/2) +(1/2)ch(u)sh(u)  =(1/2)argsh(t) +(1/2)t(√(1+t^2 )) +c  =(1/2)ln(t +(√(1+t^2 ))) +(t/2)(√(1+t^2 )) ⇒  ∫_(cotan(θ)) ^(cotan((θ/2)))  (√(1+t^2 ))dt =(1/2)[ln(t+(√(1+t^2 ))) +t(√(1+t^2 ))]_(cotanθ) ^(cotan((θ/2)))   =(1/2){ln(cotan((θ/2))+(√(1+cotan^2 ((θ/2)))))  +cotan((θ/2))(√(1+cotan^2 ((θ/2))))−ln(cotanθ+(√(1+cotanθ))  −cotanθ(√(1+cotan^2 θ)))

1)wehavef(θ)=01x2+2cosθx+1dx=01(x+cosθ)2+sin2θdθchang.x+cosθ=sinθtgivef(θ)=coatanθcotan(θ2)sinθsinθ1+t2dtf(θ)=∣sinθsinθcotan(θ)cotan(θ2)1+t2dtletfind1+t2dtchangementt=sh(u)give1+t2dt=ch(u)ch(u)du=ch2(u)du=1+ch(2u)2du=u2+14sh(2u)=u2+12ch(u)sh(u)=12argsh(t)+12t1+t2+c=12ln(t+1+t2)+t21+t2cotan(θ)cotan(θ2)1+t2dt=12[ln(t+1+t2)+t1+t2]cotanθcotan(θ2)=12{ln(cotan(θ2)+1+cotan2(θ2))+cotan(θ2)1+cotan2(θ2)ln(cotanθ+1+cotanθcotanθ1+cotan2θ)

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19

∫(√(x^2 +2xcosθ+1)) dx  ∫(√((x+cosθ)^2 +sin^2 θ)) dx  (((x+cosθ))/2)(√((x+cosθ)^2 +sin^2 θ)) +((sin^2 θ)/2)(√((x+cosθ)^2 +sin^2 θ)) +c  so answer is  ∣(((x+cosθ))/2)(√((x+cosθ)^2 +sin^2 θ)) +((sin^2 θ)/2)(√((x+cosθ)^2 +sin^2 θ)) ∣_0 ^1   [{(((1+cosθ))/2)(√((1+cosθ)^2 +sin^2 θ)) +((sin^2 θ)/2)(√((1+cosθ)^2 +sin^2 θ)) }−{((cosθ)/2)(√(cos^2 θ+sin^2 θ)) +((sin^2 θ)/2)×(√(cos^2 θ+sin^2 θ)) }]  [(((1+cosθ))/2)(√(1+2cosθ+1)) +((sin^2 θ)/2)(√(2+2cosθ)) }−{((cosθ)/2)+((sin^2 θ)/2)}]  (((1+cosθ))/2)×2cos(θ/2)+((sin^2 θ)/2)×2cos(θ/2)−((cosθ)/2)−((sin^2 θ)/2)]  check  upto this...

x2+2xcosθ+1dx(x+cosθ)2+sin2θdx(x+cosθ)2(x+cosθ)2+sin2θ+sin2θ2(x+cosθ)2+sin2θ+csoansweris(x+cosθ)2(x+cosθ)2+sin2θ+sin2θ2(x+cosθ)2+sin2θ01[{(1+cosθ)2(1+cosθ)2+sin2θ+sin2θ2(1+cosθ)2+sin2θ}{cosθ2cos2θ+sin2θ+sin2θ2×cos2θ+sin2θ}][(1+cosθ)21+2cosθ+1+sin2θ22+2cosθ}{cosθ2+sin2θ2}](1+cosθ)2×2cosθ2+sin2θ2×2cosθ2cosθ2sin2θ2]checkuptothis...

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