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Question Number 206224 by mathzup last updated on 09/Apr/24

find  ∫_0 ^1  arctan(x^5 )dx

find01arctan(x5)dx

Answered by Frix last updated on 10/Apr/24

∫tan^(−1)  x^n  dx =_(u′=1∧v=tan^(−1)  x^n ) ^(by parts)    =xtan^(−1)  x^n  −n∫(x^n /(x^(2n) +1))dx  With n=5 and x∈[0, 1] we get  (((√5)ln (1+(√5)))/2)−(((1+(√5))ln 2)/2)+((5−(√(10(5−(√5)))))/(20))π≈.151019250

tan1xndx=bypartsu=1v=tan1xn=xtan1xnnxnx2n+1dxWithn=5andx[0,1]weget5ln(1+5)2(1+5)ln22+510(55)20π.151019250

Answered by mathzup last updated on 10/Apr/24

thank you sir

thankyousir

Answered by mathzup last updated on 10/Apr/24

by parts we get  I=[x arctan(x^5 )]_0 ^1 −∫_0 ^1 x.((5x^4 )/(1+x^(10) ))dx  =(π/4)−5∫_0 ^1 (x^5 /(1+x^(10) ))dx but  ∫_0 ^1 (x^5 /(1+x^(10) ))dx=∫_0 ^1 x^5 Σ(−1)^n x^(10n)  dx  =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1  x^(10n+5) dx  =Σ_(n=0) ^∞ (−1)^n ×(1/(10n+6))  =Σ_(n=o) ^∞ (1/(10(2n)+6))−Σ_(n=0) ^∞ (1/(10(2n+1)+6))  =Σ_(n=0) ^∞ ((1/(20n+6))−(1/(20n+16)))  =10Σ_(n=0) ^∞ (1/((20n+6)(20n+16)))  =((10)/(20^2 ))Σ_(n=0) ^∞ (1/((n+(6/(20)))(n+((16)/(20)))))  =(1/(40))×((Ψ(((16)/(20)))−Ψ((6/(20))))/(((16)/(20))−(6/(20))))=(1/(20)){Ψ((4/5))−Ψ((3/(10)))}

bypartswegetI=[xarctan(x5)]0101x.5x41+x10dx=π4501x51+x10dxbut01x51+x10dx=01x5Σ(1)nx10ndx=n=0(1)n01x10n+5dx=n=0(1)n×110n+6=n=o110(2n)+6n=0110(2n+1)+6=n=0(120n+6120n+16)=10n=01(20n+6)(20n+16)=10202n=01(n+620)(n+1620)=140×Ψ(1620)Ψ(620)1620620=120{Ψ(45)Ψ(310)}

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