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Question Number 141916 by mnjuly1970 last updated on 24/May/21

       prove  that::     I:=∫_0 ^( (π/2)) arccosh(sin(x)+cos(x))dx=(π/2)ln(2) ..

provethat::I:=0π2arccosh(sin(x)+cos(x))dx=π2ln(2)..

Answered by mindispower last updated on 25/May/21

by part withe d(arccoh(x))=(dx/( (√(1−x^2 ))))  I=xarccosh(sinx+cos(x))]_0 ^(π/2) −∫_0 ^(π/2) ((x(cos(x)−sin(x)))/( (√(sin(2x)))))dx  −∫_0 ^(π/2) ((xcos(x)−xsin(x))/( (√(sin(2x)))))dx  =−(1/( (√2)))∫_0 ^(π/2) xsin^(−(1/2)) (x)cos^(1/2) (x)+(1/( (√2)))∫_0 ^(π/2) xsin^(1/2) (x)cos^(−(1/2)) (x)  (1/( (√2)))∫_0 ^∞ ((arctg(x)((√x)−(1/( (√x)))))/(1+x^2 ))dx=A  ∫_0 ^∞ ((x^s arctan(tx))/(1+x^2 )) dx=f(t),s∈]−1,1[  f′(t)=∫_0 ^∞ (x^(s+1) /((1+x^2 )(1+t^2 x^2 )))dx  ∫_(−∞) ^∞ (x^(s+1) /((1+x^2 )(1+t^2 x^2 )))dx=a=2iπ(.(i^(s+1) /(2i(1−t^2 )))+((i^(s+1) t^2 )/(t^(s+1) .(t^2 −1))).(1/(2it)))  =((πi^(s+1) )/(1−t^2 ))(1−t^(−s) )  a=f(t)+∫_0 ^∞ (((−x)^(s+1) )/((1+x^2 )(1+t^2 x^2 )))dx  =(1+(−1)^(s+1) )f(t)  =(1+e^(iπ(s+1)) )f(t)  f(t)=(π/(1−t^2 ))(1−t^(−s) ).(e^(i(π/2)(s+1)) /(1+e^(iπ(s+1)) ))  =((π(1−t^(−s) ))/(2(1−t^2 )cos((π/2)(s+1))))=f′(t)  ∫_0 ^1 f′(t)=((x^s arctan(x))/(1+x^2 ))  =(π/(2cos((π/2)(s+1))))∫_0 ^1 ((1−t^(−s) )/(1−t^2 ))dt_(t^2 =x)   =−(π/(4sin(((πs)/2))))∫_0 ^1 ((1−x^(−(s/2)) )/(1−x))x^(−(1/2))   =−(π/(4sin(((πs)/2))))∫_0 ^1 ((x^(−(1/2)) −x^(−((s+1)/2)) )/(1−x))dx=g(s)  Ψ(s+1)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx  g(s)=−(π/(4sin(((πs)/2)))).(Ψ(((1−s)/2))−Ψ((1/2)))  A=−(π/(4sin((π/4))(√2)))(Ψ((1/4))−2Ψ((1/2))+Ψ((3/4)))  Ψ((1/2))=−2ln(2)−γ  Ψ((1/4))=(−(π/2)−3ln(2)−γ)  Ψ((3/4))=Ψ((1/4))+π  =−(π/4)(−π−6ln(2)−2γ+π+4ln(2)+2γ)  =−(π/4).−2ln(2)=(π/2)ln(2)

bypartwithed(arccoh(x))=dx1x2I=xarccosh(sinx+cos(x))]0π20π2x(cos(x)sin(x))sin(2x)dx0π2xcos(x)xsin(x)sin(2x)dx=120π2xsin12(x)cos12(x)+120π2xsin12(x)cos12(x)120arctg(x)(x1x)1+x2dx=A0xsarctan(tx)1+x2dx=f(t),s]1,1[f(t)=0xs+1(1+x2)(1+t2x2)dxxs+1(1+x2)(1+t2x2)dx=a=2iπ(.is+12i(1t2)+is+1t2ts+1.(t21).12it)=πis+11t2(1ts)a=f(t)+0(x)s+1(1+x2)(1+t2x2)dx=(1+(1)s+1)f(t)=(1+eiπ(s+1))f(t)f(t)=π1t2(1ts).eiπ2(s+1)1+eiπ(s+1)=π(1ts)2(1t2)cos(π2(s+1))=f(t)01f(t)=xsarctan(x)1+x2=π2cos(π2(s+1))011ts1t2dtt2=x=π4sin(πs2)011xs21xx12=π4sin(πs2)01x12xs+121xdx=g(s)Ψ(s+1)=γ+011xs1xdxg(s)=π4sin(πs2).(Ψ(1s2)Ψ(12))A=π4sin(π4)2(Ψ(14)2Ψ(12)+Ψ(34))Ψ(12)=2ln(2)γΨ(14)=(π23ln(2)γ)Ψ(34)=Ψ(14)+π=π4(π6ln(2)2γ+π+4ln(2)+2γ)=π4.2ln(2)=π2ln(2)

Commented by mnjuly1970 last updated on 25/May/21

very nice .thanks alot...

verynice.thanksalot...

Commented by mindispower last updated on 25/May/21

withe pleasur sir

withepleasursir

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