Question

利用换元法求定积分

1、∫[√（1-x^2)/x]dx
2、∫[√(x^2-1)/x]dx
3、∫[(x-1)*e^(x^2-2x)]dx
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要有过程

Answer 1

1、令x=sint,(t属于[-π/2,0)(0,π/2]，1>cost>=0)
∫[√（1-x^2)/x]dx=∫(cost/sint)dsint
=∫(cost*cost/sint)dt
=∫[(1-sint*sint)/sint]dt
=∫(1/sint-sint)dt
=∫dt/sint+cost
=cost+∫[tan(t/2)+cot(t/2)]d(t/2)
=cost+∫[-dcos(t/2)/cos(t/2)+dsin(t/2)/sin(t/2)]
=cost-ln cos(t/2)+ln |sin(t/2)|
=cost+ln|tan(t/2)|
=根号(1-x^2)+ln|x/[1+根号(1-x^2)]|
2、令x=1/sint,t属于[-π/2,0)(0,π/2]，1>cost>=0
∫[√(x^2-1)/x]dx=∫|cott|sintd(1/sint)
=-∫(|cott|/sint)dt
=-∫{(t/|t|)[cost/(sint*sint)]}dt
=-∫{(t/|t|)[1/(sint*sint)]}dsint
=∫{(t/|t|)d(1/sint)
=∫(x/|x|)dx
=|x|
3、令t=x-1,dt=dx
∫[(x-1)*e^(x^2-2x)]dx =1/e*∫(te^t^2)dt
=1/(2e)*∫(e^t^2)dt^2
=1/2e^(t^2-1)
=1/2e^(x^2-2x)

Answer 2

请参见图片

Answer 3

那个钩钩是什么？你把题目写清点

Answer 4

确实看不清