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Calcul-pi-2-0-t-tan-t-dt-




Question Number 195895 by Erico last updated on 12/Aug/23
Calcul        ∫^( (π/2)) _( 0) t(√(tan(t))) dt
Calcul0π2ttan(t)dt
Answered by witcher3 last updated on 13/Aug/23
∫_0 ^∞ ((tan^(−1) (t).(√t))/(1+t^2 ))  =∫_0 ^∞ ((√t)/(1+t^2 )).∫_0 ^1 ((tdx)/(1+(xt)^2 )).dx  =∫_0 ^1 ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dtdx  ∫_0 ^∞ (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))dt=A  ∫_(−∞) ^0 (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=∫_0 ^∞ ((−it^(3/2) )/((1+t^2 )(1+x^2 t^2 )))dt=A  ∫_C (t^(3/2) /((1+t^2 )(1+x^2 t^2 )))=(((i)^(3/2) )/((2i)(1−x^2 ))).+(((i)^(3/2) x^2 )/(x^(3/2) (x^2 −1).2ix))  =(i^(1/2) /(2(1−x^2 )))(1−(1/( (√x))))  =((i^(3/2) π)/( (√x)(1+x)(1+(√x))))=((−πe^((i3π)/4) )/( (√x)(1+(√x))(1+x))).    ∫_0 ^1 ((π(√2))/((1+t)(1+t^2 ))).  =∫_0 ^1 ((π(√2)[1+t^2 +(1−t)(1+t))/(2(1+t)(1+t^2 )))  =(π/( (√2))){∫_0 ^1 (dt/(1+t))+∫_0 ^1 (dt/(1+t^2 ))−∫_0 ^1 ((tdt)/(1+t^2 ))}  =(π/( (√2))){ln(2)+(π/4)−(1/2)ln(2))}  =(π/( (√2)))[((ln(2))/2)+(π/4)]=∫_0 ^(π/2) x(√(tan (x)))dx
0tan1(t).t1+t2=0t1+t2.01tdx1+(xt)2.dx=010t32(1+t2)(1+x2t2)dtdx0t32(1+t2)(1+x2t2)dt=A0t32(1+t2)(1+x2t2)=0it32(1+t2)(1+x2t2)dt=ACt32(1+t2)(1+x2t2)=(i)32(2i)(1x2).+(i)32x2x32(x21).2ix=i122(1x2)(11x)=i32πx(1+x)(1+x)=πei3π4x(1+x)(1+x).01π2(1+t)(1+t2).=01π2[1+t2+(1t)(1+t)2(1+t)(1+t2)=π2{01dt1+t+01dt1+t201tdt1+t2}=π2{ln(2)+π412ln(2))}=π2[ln(2)2+π4]=0π2xtan(x)dx
Commented by Rodier97 last updated on 17/Aug/23
Hi.....i′m having trouble understanding how you did  the second line...i′m still trying to notice that   tan^(−1) (t)= ∫_0 ^1 (1/(1+(xt)^2 )) dx  but where does ′′tdx′′ come from in your term  ∫_0 ^1  ((tdx)/(1+(xt)^2 )) dx  ???  please !
Hi..imhavingtroubleunderstandinghowyoudidthesecondlineimstilltryingtonoticethattan1(t)=0111+(xt)2dxbutwheredoestdxcomefrominyourterm01tdx1+(xt)2dx???please!
Commented by witcher3 last updated on 20/Aug/23
∫_0 ^1 ((tdx)/(1+(xt)^2 ))=a;xt=y⇒dy=tdx⇒a=∫_0 ^t (dy/(1+y^2 ))=tan^(−1) (t)
01tdx1+(xt)2=a;xt=ydy=tdxa=0tdy1+y2=tan1(t)

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