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  • 計(jì)算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2

    計(jì)算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2
    數(shù)學(xué)人氣:607 ℃時(shí)間:2020-06-23 19:46:56
    優(yōu)質(zhì)解答
    積分域用x + y = π/2劃分如下,對(duì)于函數(shù)cos(x + y),區(qū)域1為正值,區(qū)域2為負(fù)值.
    ∫∫ √[1 - sin²(x + y)] dxdy
    = ∫∫ √[cos²(x + y)] dxdy
    = ∫∫ |cos(x + y)| dxdy
    = ∫(0,π/2) dx [∫(0,π/2 - x) cos(x + y) dx + ∫(π/2 - x,π/2) (- )cos(x + y) dx ]
    = ∫(0,π/2) [sin(x + y):(0,π/2 - x)] dx - ∫(0,π/2) [sin(x + y):(π/2 - x,π/2)] dx
    = ∫(0,π/2) [sin(x + π/2 - x) - sinx] dx - ∫(0,π/2) [sin(x + π/2) - sin(x + π/2 - x)] dx
    = ∫(0,π/2) (1 - sinx) dx - ∫(0,π/2) (cosx - 1) dx
    = (x + cosx - sinx + x):(0,π/2)
    = (π/2 + 0 - 1 + π/2) - (0 + 1 - 0 + 0)
    = π - 2
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