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Question Number 152276 by peter frank last updated on 27/Aug/21
Prove that  ∫_0 ^π ((xtan x)/(sec x+tan x))dx=(π^2 /2)−π
Provethat0πxtanxsecx+tanxdx=π22π
Answered by Olaf_Thorendsen last updated on 27/Aug/21
F(x) = ∫_0 ^π ((xtanx)/(secx+tanx)) dx  F(x) = ∫_0 ^π xtanx(secx−tanx) dx  F(x) = ∫_0 ^π x(((sinx)/(cos^2 x))−(1+tan^2 x)+1) dx  F(x) = [x(secx−tanx+x)]_0 ^π   −∫_0 ^π (secx−tanx+x) dx  F(x) = π(π−1)  −[ln∣secx+tanx∣+ln∣cosx∣+(x^2 /2)]_0 ^π   F(x) = π(π−1)−(π^2 /2) = (π^2 /2)−π
F(x)=0πxtanxsecx+tanxdxF(x)=0πxtanx(secxtanx)dxF(x)=0πx(sinxcos2x(1+tan2x)+1)dxF(x)=[x(secxtanx+x)]0π0π(secxtanx+x)dxF(x)=π(π1)[lnsecx+tanx+lncosx+x22]0πF(x)=π(π1)π22=π22π

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