∫xcosxdx
=∫xdsinx
=xsinx-∫sinxdx
=xsinx+cosx
∫xcosxdx
=xsinx+cosx|
=sin1+cos1-[-sin(-1)+cos(-1)]
=0
令x=2sint，dx=2costdt
x上限1、下限0，则sint上限1/2、下限0，则t上限π/6、下限0，则2t上限π/3、下限0
∫√(4-x²)dx
=2∫cost√(4-4sin²t)dt
=4∫cost√cos²tdt
=4∫cos²tdt
=2∫1+cos(2t)dt
=∫1+cos(2t)d(2t)
=2t+sin(2t)
∫√(4-x²)dx
=2t+sin(2t)|
=π/3+√3/2