定积分∫[-√2,√2]√(4-x2)dx=

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定积分∫[-√2,√2]√(4-x2)dx=
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定积分∫[-√2,√2]√(4-x2)dx=
定积分∫[-√2,√2]√(4-x2)dx=

定积分∫[-√2,√2]√(4-x2)dx=

设 x = 2 sint,则 dx = 2cost*dt,积分限变为:[-π/4, π/4]
那么积分可以变换为:
∫√[4 - 4(sint)^2] * 2cost*dt
=∫4(cost)^2 *dt
=2∫2(cost)^2 *dt
=2∫[1 + cos(2t)]*dt
=2∫dt + ∫cos(2t)*d(2t)
=2t + sin2t
=2(π/4 + π/4) + [sin(π/2) - sin(-π/2)]
=π/2 + 2