Question

高数求解定积分

Answer 1

（1）令t=1+√x，则x=(t-1)^2，dx=2(t-1)dt

原式=∫(1,2) 2(t-1)/tdt
=2∫(1,2) (1-1/t)dt
=2(t-ln|t|)|(1,2)
=2(2-ln2-1)
=2-ln4
（2）令x=sect，则dx=secttantdt
原式=∫(0,π/3) tant/sect*secttantdt
=∫(0,π/3) tan^2tdt
=∫(0,π/3) (sec^2t-1)dt
=(tant-t)|(0,π/3)
=√3-π/3
（3）令t=1+√(1+x)，则x=(t-1)^2-1=t^2-2t，dx=(2t-2)dt
原式=∫(2,3) (t^2-2t)/t*(2t-2)dt
=2∫(2,3) (t^2-3t+2)dt
=(2t^3/3-3t^2+4t)|(2,3)
=18-27+12-16/3+12-8
=5/3
（4）令x=sint，则dx=costdt
原式∫(0,π/2) cost*costdt
=(1/2)*∫(0,π/2) (1+cos2t)dt
=(1/2)*[t+(1/2)*sin2t]|(0,π/2)
=π/4