第1个回答 2015-12-20
（2）令t=tanx，x=arctant，dx=dt/(1+t^2)
原式=∫(0,√3) t/(1+t^2)dt
=(1/2)*∫(0,√3) d(1+t^2)/(1+t^2)
=(1/2)*ln|1+t^2||(0,√3)
=ln2
（3）令t=x-π/2，x=t+π/2，dx=dt
原式=∫(-π/2,π/2) √[1+cos(2t+π)]dt
=∫(-π/2,π/2) √(1-cos2t)dt
=2*∫(0,π/2) √(1-cos2t)dt
=2√2*∫(0,π/2) costdt
=2√2*sint|(0,π/2)
=2√2
（4）原式=∫(0,π/2) -cos^5xd(cosx)
=-(1/6)*cos^6x|(0,π/2)
=1/6
（5）令t=sinx，dt=cosxdx
原式=∫(0,π/4) sin^4x/cosx*cosxdx
=∫(0,π/4) sin^4xdx
=(1/4)*∫(0,π/4) (2sin^2x)^2dx
=(1/4)*∫(0,π/4) (1-cos2x)^2dx
=(1/4)*∫(0,π/4) [1-2cos2x+cos^2(2x)]dx
=(1/8)*∫(0,π/4) [2-4cos2x+1+cos4x]dx
=(1/8)*[3x-(1/8)*sin2x+(1/4)*sin4x]|(0,π/4)
=(1/8)*(3π/4-1/8)
=(6π-1)/64
（6）令x=2sint，dx=2costdt
原式=∫(0,π/4) 1/2cost*2costdt
=∫(0,π/4) dt
=t|(0,π/4)
=π/4
（7）令x=sint，dx=costdt
原式=∫(0,π/2) sin^2t*cost*costdt
=(1/4)*∫(0,π/2) (1-cos2t)(1+cos2t)dt
=(1/4)*∫(0,π/2) sin^2(2t)dt
=(1/8)*∫(0,π/2) (1-cos4t)dt
=(1/8)*[t-(1/4)*sin4t]|(0,π/2)
=π/16
（8）令t=√x-1，x=(t+1)^2，dx=2(t+1)dt
原式=∫(1,2) (t+1)/t*2(t+1)dt
=2*∫(1,2) (t+2+1/t)dt
=2*[(1/2)*t^2+2t+ln|t|]|(1,2)
=4+8+2ln2-1-4
=7+2ln2本回答被网友采纳
原式=∫(0,√3) t/(1+t^2)dt
=(1/2)*∫(0,√3) d(1+t^2)/(1+t^2)
=(1/2)*ln|1+t^2||(0,√3)
=ln2
（3）令t=x-π/2，x=t+π/2，dx=dt
原式=∫(-π/2,π/2) √[1+cos(2t+π)]dt
=∫(-π/2,π/2) √(1-cos2t)dt
=2*∫(0,π/2) √(1-cos2t)dt
=2√2*∫(0,π/2) costdt
=2√2*sint|(0,π/2)
=2√2
（4）原式=∫(0,π/2) -cos^5xd(cosx)
=-(1/6)*cos^6x|(0,π/2)
=1/6
（5）令t=sinx，dt=cosxdx
原式=∫(0,π/4) sin^4x/cosx*cosxdx
=∫(0,π/4) sin^4xdx
=(1/4)*∫(0,π/4) (2sin^2x)^2dx
=(1/4)*∫(0,π/4) (1-cos2x)^2dx
=(1/4)*∫(0,π/4) [1-2cos2x+cos^2(2x)]dx
=(1/8)*∫(0,π/4) [2-4cos2x+1+cos4x]dx
=(1/8)*[3x-(1/8)*sin2x+(1/4)*sin4x]|(0,π/4)
=(1/8)*(3π/4-1/8)
=(6π-1)/64
（6）令x=2sint，dx=2costdt
原式=∫(0,π/4) 1/2cost*2costdt
=∫(0,π/4) dt
=t|(0,π/4)
=π/4
（7）令x=sint，dx=costdt
原式=∫(0,π/2) sin^2t*cost*costdt
=(1/4)*∫(0,π/2) (1-cos2t)(1+cos2t)dt
=(1/4)*∫(0,π/2) sin^2(2t)dt
=(1/8)*∫(0,π/2) (1-cos4t)dt
=(1/8)*[t-(1/4)*sin4t]|(0,π/2)
=π/16
（8）令t=√x-1，x=(t+1)^2，dx=2(t+1)dt
原式=∫(1,2) (t+1)/t*2(t+1)dt
=2*∫(1,2) (t+2+1/t)dt
=2*[(1/2)*t^2+2t+ln|t|]|(1,2)
=4+8+2ln2-1-4
=7+2ln2本回答被网友采纳