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

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/(2cos...第二题为什么除以2cost题里只是根号没有分号啊√(4-x^2)=√[4-4(sint)^2]=√{4[1-(sint)^2]=2*cost.那应该是再乘以2cost啊 怎么变成除以了我看错了，看成除号了，设x=2sint,dx=2costdt,原式=∫[0,π/2]4(sint)^2*2cost(2cost)dt=16∫[0,π/2](sint)^2(cost)^2dt=4∫[0,π/2](sin2t)^2dt=4∫[0,π/2](sin2t)^2dt=2∫[0,π/2](1-cos4t)dt=2t[[0,π/2]-(1/2)sin4t[0,π/2]=π.