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Question Number 106583 by bemath last updated on 06/Aug/20

           @bemath@  ∫_0 ^(π/2)  ((sin x dx)/(sin x+cos x)) =?

@bemath@π/20sinxdxsinx+cosx=?

Answered by bobhans last updated on 06/Aug/20

let a = ∫_0 ^(π/2)  ((sin x dx)/(sin x+cos x))  ; ♭ = ∫_0 ^(π/2)  ((cos x dx)/(sin x+cos x))  (1) a+♭ = ∫_0 ^(π/2)  dx = (π/2)   (2) a−♭ = ∫_0 ^(π/2)  ((sin x−cos x)/(sin x+cos x)) dx = −∫_1 ^1  (du/u) = 0   [ with u = sin x+cos x ]; a = ♭  therefore a = ∫_0 ^(π/2)  ((sin x dx)/(sin x+cos x)) = (1/2)×(π/2)=(π/4)

leta=π20sinxdxsinx+cosx;=π20cosxdxsinx+cosx(1)a+=π20dx=π2(2)a=π20sinxcosxsinx+cosxdx=11duu=0[withu=sinx+cosx];a=thereforea=π20sinxdxsinx+cosx=12×π2=π4

Commented by john santu last updated on 06/Aug/20

nice & cooll..

nice&cooll..

Commented by bemath last updated on 06/Aug/20

creative...⋰

creative...\iddots

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