求解：对1/2*sin(x)*x^(-1/2)的不定积分？？？
2个回答
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(1/2)∫sinxdx/x^(1/2)
=∫sinxdx^(1/2)
=sinx*x^(1/2)-∫cosx*x^(1/2)dx
cosx=1-x^2/2!+x^4/4!-x^6/6!+x^8/8!+..+(-1)^n x^2n/2n!
=sinx*x^(1/2)-[∫x^(-1/2)-x^(3/2)/2!+x^(7/2)/4!-x^(11/2)/6!+..+(-1)^nx^(2n-1/2)/2n!]
=sinx*x^(1/2)-2x^(1/2)+(2/5)x^(5/2)/2!-(2/9)x^(9/2)/4!+(2/13)x^(13/2)/6!-..-(-1)^n(2/(4n+1))x^(2n+1/2)/2n!
=∫sinxdx^(1/2)
=sinx*x^(1/2)-∫cosx*x^(1/2)dx
cosx=1-x^2/2!+x^4/4!-x^6/6!+x^8/8!+..+(-1)^n x^2n/2n!
=sinx*x^(1/2)-[∫x^(-1/2)-x^(3/2)/2!+x^(7/2)/4!-x^(11/2)/6!+..+(-1)^nx^(2n-1/2)/2n!]
=sinx*x^(1/2)-2x^(1/2)+(2/5)x^(5/2)/2!-(2/9)x^(9/2)/4!+(2/13)x^(13/2)/6!-..-(-1)^n(2/(4n+1))x^(2n+1/2)/2n!
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令x^(-1/2)=t,x=t^2,dx=2tdt
∫1/2*sin(x)*x^(-1/2)dx
=∫1/2*sint^2/t*tdt
=∫sint^2dt
不可积
∫1/2*sin(x)*x^(-1/2)dx
=∫1/2*sint^2/t*tdt
=∫sint^2dt
不可积
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