Question

用换元积分法求定积分 过程要详细

Answer 1

1、令x=2sint，则dx=2cost dt
x=0时，t=0；x=2时，t=π/3
故原式=∫[0,π/2]4sin²t · 2cost · 2cost dt
=16∫[0,π/2]sin²t cos²t dt
=4∫[0,π/2]sin²2t dt
=2∫[0,π/2](1-cos4t)dt
=2(t-1/4 sin4t)|[0,π/2]
=2(π/2 -0)
=π
2、令√(1-x²)=t，则x²=1-t²，d(x²)=-2t dt
x=0时，t=1；x=1时，t=0
故原式=½∫[0,1]√(1-x²)/(2-x²) d(x²)
=½∫[1,0]t/(1+t²) · (-2t)dt
=∫[0,1]t²dt/(1+t²)
=∫[0,1][1-1/(1+t²)]dt
=[t-arctant]|[0,1]
=1-π/4

Answer 2

如图

Answer 3

Answer 4