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0-1-arc-tan-x-2-dx-




Question Number 117403 by bemath last updated on 11/Oct/20
   ∫_0 ^1  (arc tan x)^2  dx =?
10(arctanx)2dx=?
Commented by MJS_new last updated on 11/Oct/20
use arctan x =((ln (1+ix) −ln (1−ix))/(2i)) ⇒  −(1/4)∫_0 ^1 (ln^2  (1+ix) −2ln (1+ix) ln (1−ix) +ln^2  (1−ix))dx  and these are solveable (by parts)
usearctanx=ln(1+ix)ln(1ix)2i1410(ln2(1+ix)2ln(1+ix)ln(1ix)+ln2(1ix))dxandthesearesolveable(byparts)
Answered by mindispower last updated on 11/Oct/20
=[xarctan^2 (x)]_0 ^1 −∫_0 ^1 ((2x)/(1+x^2 ))arctan(x)dx  =(π^2 /(16))−I  let arctan(x)=t  I=−∫_0 ^(π/4) ((2tg(t)t)/(1+tg^2 (t))).(1+tg^2 (t))dt  =−2∫_0 ^(π/4) tg(t)tdt by part=−2([−tln(cost)]_0 ^(π/4)   +∫_0 ^(π/4) ln(cos(t))dt  =−(π/4)ln(2)−2∫_0 ^(π/4) ln(cos(t))dt  one way too find it  let a=∫_0 ^(π/4) ln(cos(t))dt  =∫_0 ^(π/4) (1/2)ln(sin(t)cos(t).(1/(tg(t))))dt  =∫_0 ^(π/4) ((ln(((sin(2t))/2)))/2)dt−(1/2)∫_0 ^(π/4) ln(tg(t))dt_(tg(t)=s)   2t=u⇒(1/4)∫_0 ^(π/2) ln(u)du−∫_0 ^(π/4) ln(2)du in first  ∫_0 ^(π/4) ln(tg(t))dt=∫_0 ^1 ((ln(s))/(1+s^2 ))ds=Σ∫_0 ^1 (−1)^k s^(2k+1) ln(s)ds  =−Σ(−1)^k .(1/((2k+1)^2 ))=−β(2) =−G ,G catalan constante  knowing∫_0 ^(π/2) ln(sin(t))dt give us close forme  2nd use fourier serie of   ln(cos(t))=−ln(2)−Σ_(n≥1) ((cos(2(2n−1)t))/(2n−1))  give use resulte imediatly  withe Σ_(n≥1) (((−1)^(n−1)  )/((2n−1)^2 )) =β(2)
=[xarctan2(x)]01012x1+x2arctan(x)dx=π216Iletarctan(x)=tI=0π42tg(t)t1+tg2(t).(1+tg2(t))dt=20π4tg(t)tdtbypart=2([tln(cost)]0π4+0π4ln(cos(t))dt=π4ln(2)20π4ln(cos(t))dtonewaytoofinditleta=0π4ln(cos(t))dt=0π412ln(sin(t)cos(t).1tg(t))dt=0π4ln(sin(2t)2)2dt120π4ln(tg(t))dttg(t)=s2t=u140π2ln(u)du0π4ln(2)duinfirst0π4ln(tg(t))dt=01ln(s)1+s2ds=Σ01(1)ks2k+1ln(s)ds=Σ(1)k.1(2k+1)2=β(2)=G,Gcatalanconstanteknowing0π2ln(sin(t))dtgiveuscloseforme2ndusefourierserieofln(cos(t))=ln(2)n1cos(2(2n1)t)2n1giveuseresulteimediatlywithen1(1)n1(2n1)2=β(2)

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