求定积分 ∫[0,1] (1+x^2)^(-3/2) dx; ∫[0,2] x^2√(4-x^2)

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1、设x=tant,dx=(sect)^2dt,
原式= ∫[0,π/4][（sect)^2]^(-3/2)*(sect)^2dt
=∫[0,π/4](sect)^2dt/(sect)^3
=∫[0,π/4]costdt
=sint[0,π/4]
=√2/2.
2、设x=2sint,dx=2costdt,
原式=∫[0,π/2]4(sint)^2*2costdt/(2cost)
=4∫[0,π/2](sint)^2dt
=2∫[0,π/2](1-cos2t)dt
=(2t)[0,π/2]-(sin2t))[0,π/2]
=π.