这个定积分怎么算呢？求详细过程
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用换元积分法求解
令x=√2*sint，则dx=√2*costdt
原式=(4/3)*∫(0,π/4) (2-2sin^2t)^(3/2)*√2*costdt
=(4/3)*2^(3/2)*√2*∫(0,π/4) cos^4tdt
=(16/3)*∫(0,π/4) (1/4)*(2cos^2t)^2dt
=(4/3)*∫(0,π/4) (1+cos2t)^2dt
=(4/3)*∫(0,π/4) [1+2cos2t+(cos2t)^2]dt
=(4/3)*∫(0,π/4) [1+2cos2t+(1/2)*(1+cos4t)]dt
=(4/3)*∫(0,π/4) [3/2+2cos2t+(1/2)*cos4t]dt
=(4/3)*[3t/2+sin2t+(1/8)*sin4t]|(0,π/4)
=(4/3)*(3π/8+1)
=π/2+4/3
令x=√2*sint，则dx=√2*costdt
原式=(4/3)*∫(0,π/4) (2-2sin^2t)^(3/2)*√2*costdt
=(4/3)*2^(3/2)*√2*∫(0,π/4) cos^4tdt
=(16/3)*∫(0,π/4) (1/4)*(2cos^2t)^2dt
=(4/3)*∫(0,π/4) (1+cos2t)^2dt
=(4/3)*∫(0,π/4) [1+2cos2t+(cos2t)^2]dt
=(4/3)*∫(0,π/4) [1+2cos2t+(1/2)*(1+cos4t)]dt
=(4/3)*∫(0,π/4) [3/2+2cos2t+(1/2)*cos4t]dt
=(4/3)*[3t/2+sin2t+(1/8)*sin4t]|(0,π/4)
=(4/3)*(3π/8+1)
=π/2+4/3
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