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Question Number 98623 by hardylanes last updated on 15/Jun/20

evaluate   ∫_(2/(√3)) ^2 (1/(x^2 (√(4+x^2 ))))dx using the substitution x=2tanθ

evaluate2321x24+x2dxusingthesubstitutionx=2tanθ

Answered by Kunal12588 last updated on 15/Jun/20

x=2tanθ  ⇒dx=2sec^2 θ dθ  θ=tan^(−1) ((x/2));   ∫_( π/6) ^( π/4) ((2sec^2 θ)/(4tan^2 θ(√(4+4tan^2 θ))))dθ  ∫_( π/6) ^( π/4) ((secθ)/(4tan^2 θ))dθ  (1/4)∫_( π/6) ^( π/4) ((cosθ)/(sin^2 θ))dθ  (1/4)∫_( π/6) ^( π/4) ((d(sinθ))/(sin^2 θ))  (1/4)[−(1/(sin θ))]_(π/6) ^(π/4) =(1/4)(−(√2)+2)=((2−(√2))/4)

x=2tanθdx=2sec2θdθθ=tan1(x2);π/6π/42sec2θ4tan2θ4+4tan2θdθπ/6π/4secθ4tan2θdθ14π/6π/4cosθsin2θdθ14π/6π/4d(sinθ)sin2θ14[1sinθ]π/6π/4=14(2+2)=224

Commented by hardylanes last updated on 15/Jun/20

thanks

Commented by abdomathmax last updated on 15/Jun/20

your answer is correct miss kunal

youransweriscorrectmisskunal

Commented by Kunal12588 last updated on 16/Jun/20

it′s mister!

itsmister!

Answered by MWSuSon last updated on 15/Jun/20

x=2tanθ, dx=2sec^2 xdθ  (2/(√3))=2tanθ, θ=arctan((1/(√3)))=(π/6)  2=2tanθ, θ=arctan(1)=(π/4)  ∫_(π/6) ^(π/4) ((2sec^2 θ)/(4tan^2 θ(√(4+4tan^2 θ))))dθ=∫_(π/6) ^(π/4) ((sec^2 θ)/(2×2tan^2 θ(√(1+tan^2 θ))))dθ  =∫_(π/6) ^(π/4) ((secθ)/(4tan^2 θ))dθ=(1/4)∫_(π/6) ^(π/4) (1/(cosθ))×((cos^2 θ)/(sin^2 θ))dθ=(1/4)∫_(π/6) ^(π/4) ((cosθ)/(sin^2 θ))dθ  let t=sinθ, dt=cosθdθ  =(1/4)∫_(π/6) ^(π/4) (dt/t^2 )=(1/4)(−(1/(sin((π/4))))+(1/(sin((π/6)))))=(1/4)(2−(√2))

x=2tanθ,dx=2sec2xdθ23=2tanθ,θ=arctan(13)=π62=2tanθ,θ=arctan(1)=π4π4π62sec2θ4tan2θ4+4tan2θdθ=π4π6sec2θ2×2tan2θ1+tan2θdθ=π4π6secθ4tan2θdθ=14π4π61cosθ×cos2θsin2θdθ=14π4π6cosθsin2θdθlett=sinθ,dt=cosθdθ=14π4π6dtt2=14(1sin(π4)+1sin(π6))=14(22)

Commented by hardylanes last updated on 15/Jun/20

thankiu

Answered by abdomathmax last updated on 15/Jun/20

let I =∫_(2/(√3)) ^2  (dx/(x^2 (√(4+x^2 ))))  let do the changement   x =2sht ⇒t =argsh((x/2)) ⇒  I = ∫_(ln((1/(√3))+(√(1+(1/3))))) ^(ln(1+(√2)))  ((2ch(t)dt)/(4sh^2 t(2cht)))    = ∫_(ln((√3))) ^(ln(1+(√2)))   (dt/(4(((ch(2t)−1)/2))))  =(1/2) ∫_(ln((√3))) ^(ln(1+(√2)))   (dt/(((e^(2t) +e^(−2t) )/2)−1))  =∫_(ln((√3))) ^(ln(1+(√2)))    (dt/(e^(2t)  +e^(−2t) −2))  =_(e^(2t)  =u)     ∫_3 ^((1+(√2))^2 )     (du/(2u(u +u^(−1) −2)))  =∫_3 ^(3+2(√2))      (du/(2(u^2 +1−2u)))  =(1/2) ∫_3 ^(3+2(√2))   (du/((u−1)^2 )) =(1/2)[−(1/(u−1))]_3 ^(3+2(√2))   =(1/2){(1/2)−(1/(2+2(√2)))} =(1/4){1−(1/(1+(√2)))}  =(1/4){((√2)/(1+(√2)))} =(1/4)×(((√2)((√2)−1))/1) =(1/4)(2−(√2))  =(1/2)−((√2)/4)

letI=232dxx24+x2letdothechangementx=2shtt=argsh(x2)I=ln(13+1+13)ln(1+2)2ch(t)dt4sh2t(2cht)=ln(3)ln(1+2)dt4(ch(2t)12)=12ln(3)ln(1+2)dte2t+e2t21=ln(3)ln(1+2)dte2t+e2t2=e2t=u3(1+2)2du2u(u+u12)=33+22du2(u2+12u)=1233+22du(u1)2=12[1u1]33+22=12{1212+22}=14{111+2}=14{21+2}=14×2(21)1=14(22)=1224

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