计算:定积分∫(在上√2 ,在下 0)(√2-X^2) dx求详细过程答案,拜托大神...

如题所述

解:
令x=√2·sint,则dx=√2·costdt
∫(0→√2)√(2-x²)dx
=∫(0→π/2)√2·cost·√2·cost dt
=∫(0→π/2)2·cos²t dt
=∫(0→π/2)(1+cos2t) dt
=[t+1/2·sin2t]|(0→π/2)
=π/2
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