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Question Number 71490 by oyemi kemewari last updated on 16/Oct/19

Commented by mathmax by abdo last updated on 16/Oct/19

let A=∫_0 ^(1/2)  ((1+(√3))/((^4 (√((1+x)^2 (1−x)^6 ))))dx ⇒A=(1+(√3))∫_0 ^(1/2)   (dx/((√(1+x))(1−x)^(2/3) ))  changement x=cos(2t) give t=(1/2)arcosx  A=(1+(√3))∫_(π/4) ^(π/6)    ((−2sin2t dt)/((√(2cos^2 t))(2sin^2 t)^(3/2) ))  =((2(1+(√3)))/2^((1/2)+(3/2)) ) ∫_(π/6) ^(π/4)    ((sin(2t))/(cost(sint)(sint)^2 ))dt  =((2(1+(√3)))/2^2 ) ∫_(π/6) ^(π/4)  ((2sint cost)/(cost sint (sint)^2 ))dt=(1+(√3))∫_(π/6) ^(π/4)   (dt/(sin^2 t))  =(1+(√3))∫_(π/6) ^(π/4)    (dt/((1−cos(2t))/2)) =2(1+(√3)) ∫_(π/6) ^(π/4)    (dt/(1−cos(2t)))  but ∫_(π/6) ^(π/4)   (dt/(1−cos(2t))) =_(tant=u)    ∫_(1/(√3)) ^1    (1/(1−((1−u^2 )/(1+u^2 )))) (du/(1+u^2 ))  =∫_(1/(√3)) ^1  (du/(1+u^2 −1+u^2 )) =∫_(1/(√3)) ^1   (du/(2u^2 )) =(1/2)[−(1/u)]_(1/(√3)) ^1   =(1/2)((√3)−1) ⇒A=2(1+(√3))×(((√3)−1)/2) =((√3))^2 −1 =2 ⇒  A =2

letA=0121+3(4(1+x)2(1x)6dxA=(1+3)012dx1+x(1x)23changementx=cos(2t)givet=12arcosxA=(1+3)π4π62sin2tdt2cos2t(2sin2t)32=2(1+3)212+32π6π4sin(2t)cost(sint)(sint)2dt=2(1+3)22π6π42sintcostcostsint(sint)2dt=(1+3)π6π4dtsin2t=(1+3)π6π4dt1cos(2t)2=2(1+3)π6π4dt1cos(2t)butπ6π4dt1cos(2t)=tant=u131111u21+u2du1+u2=131du1+u21+u2=131du2u2=12[1u]131=12(31)A=2(1+3)×312=(3)21=2A=2

Commented by mathmax by abdo last updated on 16/Oct/19

error of typo at first line A=(1+(√3))∫_0 ^(1/2)   (dx/((√(1+x))(1−x)^(3/2) ))

erroroftypoatfirstlineA=(1+3)012dx1+x(1x)32

Answered by mind is power last updated on 16/Oct/19

x=sin(α)  ∫_0 ^(π/6) ((1+(√3))/(((1+sin(α))^2 (1−sin(α))^6 ))^(1/4) )cos(α)dα  1+_− sin(α)=(cos((α/2))+_− sin((α/2)))^2   ∫_0 ^(π/6) ((1+(√3))/(((1+sin(α))^2 (1−sin(α))^6 ))^(1/4) )cos(α)dα=∫_0 ^(π/6) (((1+(√3))cos(α)dα)/(((cos((α/2))+sin((α/2)))^4 (cos((α/2))−sin((α/2)))^(12) ))^(1/4) )  =∫_0 ^(π/6) (((1+(√3))cos(α))/((cos((α/2))+sin((α/2))).(cos((α/2))−sin((α/2)))^3 ))  =∫_0 ^(π/6) (((1+(√3))cos(α)dα)/(cos(α).(cos((α/2))−sin((α/2)))^2 ))  =∫_0 ^(π/6) ((1+(√3))/(cos^2 ((α/2))(1−tg((α/2)))^2 ))dα  =(1+(√3))∫_0 ^(π/6) ((1+tg^2 ((α/2)))/((1−tg((α/2)))^2 ))dα  =2(1+(√3)).[(1/(1−tg((α/2))))]_0 ^(π/6) =2(1+(√3)).(1/(1−tg((π/(12)))))−2−2(√3)=((2(1+(√3))(((1+cos((π/6))+sin((π/6)))/2)))/(cos((π/6))))−2−2(√3)  =(((1+(√3))(1+((√3)/2)+(1/2)))/((√3)/2))=(((3+(√3))(1+(√3)))/(√3)) −2−2(√3)=4+2(√3)−2−2(√3)=2

x=sin(α)0π61+3(1+sin(α))2(1sin(α))64cos(α)dα1+sin(α)=(cos(α2)+sin(α2))20π61+3(1+sin(α))2(1sin(α))64cos(α)dα=0π6(1+3)cos(α)dα(cos(α2)+sin(α2))4(cos(α2)sin(α2))124=0π6(1+3)cos(α)(cos(α2)+sin(α2)).(cos(α2)sin(α2))3=0π6(1+3)cos(α)dαcos(α).(cos(α2)sin(α2))2=0π61+3cos2(α2)(1tg(α2))2dα=(1+3)0π61+tg2(α2)(1tg(α2))2dα=2(1+3).[11tg(α2)]0π6=2(1+3).11tg(π12)223=2(1+3)(1+cos(π6)+sin(π6)2)cos(π6)223=(1+3)(1+32+12)32=(3+3)(1+3)3223=4+23223=2

Commented by MJS last updated on 16/Oct/19

I guess something went wrong...

Iguesssomethingwentwrong...

Commented by mind is power last updated on 16/Oct/19

yeah   [f(x)]_0 ^(π/2) =f((π/2))−f(0)  i firget too substract f(0)

yeah[f(x)]0π2=f(π2)f(0)ifirgettoosubstractf(0)

Commented by mathmax by abdo last updated on 16/Oct/19

the answer is correct sir mjs.

theansweriscorrectsirmjs.

Answered by MJS last updated on 16/Oct/19

∫((1+(√3))/(((1+x)^2 (1−x)^6 ))^(1/4) )dx=  =(1+(√3))∫(dx/(√(∣x−1∣^3 ∣x+1∣)))  0≤x≤(1/2) ⇒ (1/(√(∣x−1∣^3 ∣x+1∣)))=(1/(√((−x+1)^3 (x+1))))  ∫(dx/(√((−x+1)^3 (x+1))))=       [t=arcsin (√((−x+1)/2)) → dx=−2(√((−x+1)(x+1)))]  =−∫(dt/(sin^2  t))=(1/(tan t))=((√(x+1))/(√(−x+1)))+C  ∫_0 ^(1/2) ((1+(√3))/(((1+x)^2 (1−x)^6 ))^(1/4) )dx=2

1+3(1+x)2(1x)64dx==(1+3)dxx13x+10x121x13x+1=1(x+1)3(x+1)dx(x+1)3(x+1)=[t=arcsinx+12dx=2(x+1)(x+1)]=dtsin2t=1tant=x+1x+1+C1/201+3(1+x)2(1x)64dx=2

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