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Question Number 75950 by turbo msup by abdo last updated on 21/Dec/19
give ∫_0 ^(π/2) (x/(sinx))dx at form of serie.
give∫0π2xsinxdxatformofserie.
Commented by mathmax by abdo last updated on 22/Dec/19
let A = ∫_0 ^(π/2) (x/(sinx))dx changement tan((x/2))=t give A =2∫_0 ^1 ((arctan(t))/((2t)/(1+t^2 )))((2dt)/(1+t^2 )) =2 ∫_0 ^1 ((arctan(t))/t)dt we have (d/dt)(arctant) =(1/(1+t^2 )) =Σ_(n=0) ^∞ (−1)^n t^(2n) ⇒ arctan(t) =Σ_(n=0) ^∞ (−1)^n (t^(2n+1) /(2n+1)) +c (c=0) ⇒ ((arctan(t))/t) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))t^(2n) ⇒∫_0 ^1 ((arctan(t))/t)dt=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))∫_0 ^1 t^(2n) dt =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))[(1/(2n+1))t^(2n+1) ]_0 ^1 =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) ⇒ A =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))
letA=∫0π2xsinxdxchangementtan(x2)=tgiveA=2∫01arctan(t)2t1+t22dt1+t2=2∫01arctan(t)tdtwehaveddt(arctant)=11+t2=∑n=0∞(−1)nt2n⇒arctan(t)=∑n=0∞(−1)nt2n+12n+1+c(c=0)⇒arctan(t)t=∑n=0∞(−1)n2n+1t2n⇒∫01arctan(t)tdt=∑n=0∞(−1)n(2n+1)∫01t2ndt=∑n=0∞(−1)n(2n+1)[12n+1t2n+1]01=∑n=0∞(−1)n(2n+1)2⇒A=∑n=0∞(−1)n(2n+1)2
Commented by mathmax by abdo last updated on 22/Dec/19
forgive A =2 Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))
forgiveA=2∑n=0∞(−1)n(2n+1)2

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