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Question Number 134420 by liberty last updated on 03/Mar/21

∫^( π/2) _(−π/2) (1/(2019^x +1)). ((sin^(2020) x)/(sin^(2020) x+cos^(2020) x)) dx ?

∫−π/2π/212019x+1.sin2020xsin2020x+cos2020xdx?

Answered by Dwaipayan Shikari last updated on 03/Mar/21

∫_(−(π/2)) ^(π/2) (1/(2019^x +1)).((sin^(2020) x)/(sin^(2020) x+cos^(2020) x))dx=∫_(−(π/2)) ^(π/2) ((2019^x )/(2019^x +1)).((sin^(2020) x)/(sin^(2020) x+cos^(2020) x))dx ⇒2I=∫_(−(π/2)) ^(π/2) ((sin^(2020) x)/(sin^(2020) x+cos^(2020) x))dx ⇒I=∫_0 ^(π/2) ((sin^(2020) x)/(sin^(2020) x+cos^(2020) x))dx=∫_0 ^(π/2) ((cos^(2020) x)/(sin^(2020) x+cos^(2020) x))dx 2I=∫_0 ^(π/2) 1dx⇒I=(π/4)

∫−π2π212019x+1.sin2020xsin2020x+cos2020xdx=∫−π2π22019x2019x+1.sin2020xsin2020x+cos2020xdx⇒2I=∫−π2π2sin2020xsin2020x+cos2020xdx⇒I=∫0π2sin2020xsin2020x+cos2020xdx=∫0π2cos2020xsin2020x+cos2020xdx2I=∫0π21dx⇒I=π4

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