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Question Number 104264 by mohammad17 last updated on 20/Jul/20

∫_0 ^( (√2))  ∫_y ^( (√(4−y^2 ))) (1/(√(1+x^2 +y^2 )))dxdy

02y4y211+x2+y2dxdy

Answered by Ar Brandon last updated on 20/Jul/20

I=∫_0 ^(√2) {∫_y ^(√(4−y^2 )) (dx/(√(1+x^2 +y^2 )))}dy=∫_0 ^(√2) (1/(√(1+y^2 )))∙{∫_y ^(√(4−y^2 )) (dx/(√(1+((x/(√(1+y^2 ))))^2 )))}dy     =∫_0 ^(√2) (1/(√(1+y^2 )))∙((√(1+y^2 ))/1){Arcsinh((x/(√(1+y^2 ))))}_y ^(√(4−y^2 )) dy     =∫_0 ^(√2) {Arcsinh((√((4−y^2 )/(1+y^2 ))))−Arcsinh((y/(√(1+y^2 ))))}dy  To be continued...

I=02{y4y2dx1+x2+y2}dy=0211+y2{y4y2dx1+(x1+y2)2}dy=0211+y21+y21{Arcsinh(x1+y2)}y4y2dy=02{Arcsinh(4y21+y2)Arcsinh(y1+y2)}dyTobecontinued...

Commented by mohammad17 last updated on 20/Jul/20

sir by polar

sirbypolar

Answered by Ar Brandon last updated on 21/Jul/20

  En coordonne^� es polaires x=rcosθ , y=rsinθ  0≤y≤(√2) ⇒0≤rsinθ≤(√2) ⇒0≤r≤((√2)/(sinθ))  y≤x≤(√(4−y^2 )) ⇒ { ((y≤x )),((x≤(√(4−y^2 )))) :}⇒ { ((tanθ≤1)),((x^2 +y^2 ≤4)) :}⇒ { ((θ≤(π/4))),((r≤2)) :}  r=2⇒0≤2sinθ≤(√2) ⇒0≤θ≤(π/4)  ⇒I=∫_0 ^(π/4) ∫_0 ^((√2)/(sinθ)) (r/(√(1+r^2 )))drdθ=∫_0 ^(π/4) [(√(1+r^2 ))]_0 ^((√2)/(sinθ)) dθ  ⇒I=∫_0 ^(π/4) {(√(1+(2/(sin^2 θ))))−1}dθ=∫_0 ^(π/4) {(√(1+(2/(sin^2 θ))))}dθ−(π/4)              A^�  suivre...

Encoordonnees´polairesx=rcosθ,y=rsinθ0y20rsinθ20r2sinθyx4y2{yxx4y2{tanθ1x2+y24{θπ4r2r=202sinθ20θπ4I=0π402sinθr1+r2drdθ=0π4[1+r2]02sinθdθI=0π4{1+2sin2θ1}dθ=0π4{1+2sin2θ}dθπ4A`suivre...

Commented by Ar Brandon last updated on 21/Jul/20

  I =∫_0 ^(π/4)  ((√(sin^2 θ+2))/(sinθ))dθ ⇒I =∫_0 ^(π/4)  ((sinθ(√(1+(2/(sin^2 θ)))))/(sinθ))dθ =∫_0 ^(π/4) (√(1+(2/(1−cos^2 θ))))dθ  we have 1+tan^2 θ =(1/(cos^2 θ)) ⇒cos^2 θ =(1/(1+tan^2 θ)) ⇒1−cos^2 θ=1−(1/(1+tan^2 θ))  =((tan^2 θ)/(1+tan^2 θ)) ⇒(2/(1−cos^2 θ)) =2×((1+tan^2 θ)/(tan^2 θ)) ⇒  I =∫_0 ^(π/4) (√(1+((2+2tan^2 θ)/(tan^2 θ))))dθ  =∫_0 ^(π/4) (√((3tan^2 θ+2)/(tan^2 θ)))dθ  =_(tanθ =x)   =∫_0 ^1 (√((3x^2 +2)/x^2 )) (dx/(1+x^2 ))  =∫_0 ^1   ((√(3x^2 +2))/(x(1+x^2 )))dx  =(√3)∫_0 ^1  ((√(x^2  +(2/3)))/(x(1+x^2 )))dx  cha7gement x =(√(2/3))shu give u =argsh((√(3/2))x)  I =(√3)∫_0 ^(arhsh((√(3/2))))   (2/3)×((chu)/((√(2/3))sh(u)(1+(2/3)sh^2 u)))×(√(2/3))ch(u)du  =2(√3)∫_0 ^(ln((√(3/2))+(√(1+(9/4)))))    ((ch^2 u)/(shu(3+2sh^2 u)))du    and this integral  can be solved  ...be continued...

I=0π4sin2θ+2sinθdθI=0π4sinθ1+2sin2θsinθdθ=0π41+21cos2θdθwehave1+tan2θ=1cos2θcos2θ=11+tan2θ1cos2θ=111+tan2θ=tan2θ1+tan2θ21cos2θ=2×1+tan2θtan2θI=0π41+2+2tan2θtan2θdθ=0π43tan2θ+2tan2θdθ=tanθ=x=013x2+2x2dx1+x2=013x2+2x(1+x2)dx=301x2+23x(1+x2)dxcha7gementx=23shugiveu=argsh(32x)I=30arhsh(32)23×chu23sh(u)(1+23sh2u)×23ch(u)du=230ln(32+1+94)ch2ushu(3+2sh2u)duandthisintegralcanbesolved...becontinued...

Commented by Ar Brandon last updated on 21/Jul/20

  ((ch^2 u)/(shu(3+2sh^2 u)))=((1+sh^2 u)/(shu(3+2sh^2 u)))  ((1+t^2 )/(t(3+2t^2 )))=((at+b)/(2t^2 +3))+(c/t)=(((at+b)t+c(2t^2 +3))/(t(2t^2 +3)))  t→0 , c=(1/3) , a+2c=1, a=(1/3) , b=0  ⇒((1+sh^2 u)/(shu(3+2sh^2 u)))=(((1/3)shu)/(3+2sh^2 u))+((1/3)/(shu))  ⇒∫((ch^2 u)/(shu(3+2sh^2 u)))du=(1/3){∫((shu du)/(3+2ch^2 −2))+∫(du/(shu))}  =(1/3){(1/(√2))∫((d((√2)chu))/(1+2ch^2 u))+∫((shu du)/(ch^2 u−1))}  =(1/3){(1/(√2))Arctan((√2)chu)−Arctanh(chu)}

ch2ushu(3+2sh2u)=1+sh2ushu(3+2sh2u)1+t2t(3+2t2)=at+b2t2+3+ct=(at+b)t+c(2t2+3)t(2t2+3)t0,c=13,a+2c=1,a=13,b=01+sh2ushu(3+2sh2u)=13shu3+2sh2u+13shuch2ushu(3+2sh2u)du=13{shudu3+2ch22+dushu}=13{12d(2chu)1+2ch2u+shuduch2u1}=13{12Arctan(2chu)Arctanh(chu)}

Commented by Ar Brandon last updated on 21/Jul/20

With mr mathmax′s help

Withmrmathmaxshelp

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