高等数学,三角函数求积分,
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=8/15 + (3π)/16
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I = (1/4)∫<0,π/2>(1-cos2t)^2dt - ∫<0,π/2>[1-(cost)^2]^2dcost
= (1/4)∫<0,π/2>[3/2-2cos2t+(1/2)cos4t]dt
- ∫<0,π/2>[1-2(cost)^2+(cost)^4]dcost
= (1/4)[3t/2-sin2t+(1/8)sin2t])<0,π/2>
- [cost-(2/3)(cost)^3+(1/5)(cost)^5]<0,π/2>
= 3π/16 +(1-2/3+1/5) = 3π/16 + 8/15 .
= (1/4)∫<0,π/2>[3/2-2cos2t+(1/2)cos4t]dt
- ∫<0,π/2>[1-2(cost)^2+(cost)^4]dcost
= (1/4)[3t/2-sin2t+(1/8)sin2t])<0,π/2>
- [cost-(2/3)(cost)^3+(1/5)(cost)^5]<0,π/2>
= 3π/16 +(1-2/3+1/5) = 3π/16 + 8/15 .
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