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Question Number 138114 by mnjuly1970 last updated on 10/Apr/21

            ...... calculus.....(III)......           evaluate::                  𝛗=^(???) ∫_0 ^( (π/2)) ∫_0 ^( x) ((cos(y))/( (√(((π/2)−x)((π/2)−y)))))dydx

......calculus.....(III)......evaluate::ϕ=???0π20xcos(y)(π2x)(π2y)dydx

Commented by Kamel last updated on 10/Apr/21

Commented by mnjuly1970 last updated on 10/Apr/21

thanks alot mr..kamel

thanksalotmr..kamel

Answered by mnjuly1970 last updated on 10/Apr/21

  𝛗=∫_0 ^( (π/2)) ∫_y ^( (π/2)) ((cos(y))/( (√(((π/2)−x)((π/2)−y)))))dxdy  =2∫_0 ^( (π/2)) ((cos(y))/( (√((π/2)−y)))) [−(√((π/2)−x)) ]_y ^(π/2) dy  =2∫_0 ^(π/2) cos(y)dy=2sin((π/2))=2...✓

ϕ=0π2yπ2cos(y)(π2x)(π2y)dxdy=20π2cos(y)π2y[π2x]yπ2dy=20π2cos(y)dy=2sin(π2)=2...

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