∫sinx/(1+sinx) dx=?如题,做了好久做不出,大家帮做下,
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∫sinx/(1+sinx) dx=?
如题,做了好久做不出,大家帮做下,
原式=∫sinxdx+∫(1-cos2x)/2dx
=-cos x+1/2*x-1/2∫cos2xdx
=-cosx+1/2*x-1/4sin2x+C
∫[(sinx+1)-1]/(sinx+1)dx=∫[1-1/(1+sinx)]dx=x+c-∫1/(1+sinx)dx
(ctanx)'=(cosx/sinx)'=-1/sin^2(x) (1)
1/(1+sinx)=1/[1+2sin(x/2)cos(x/2)]=1/[sin(x/2)+cos(x/2)]^2=1/[根号(2)sin(x/2+45º)]^2=1/2[...
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∫[(sinx+1)-1]/(sinx+1)dx=∫[1-1/(1+sinx)]dx=x+c-∫1/(1+sinx)dx
(ctanx)'=(cosx/sinx)'=-1/sin^2(x) (1)
1/(1+sinx)=1/[1+2sin(x/2)cos(x/2)]=1/[sin(x/2)+cos(x/2)]^2=1/[根号(2)sin(x/2+45º)]^2=1/2[sin(x/2+45º)]^2
则∫1/(1+sinx)dx=∫1/2[sin(x/2+45º)]^2 dx=∫1/[sin(x/2+45º)]^2 d(x/2+45º)
根据(1),令y=x/2+45º,则上式变为∫1/sin^2(y) dy=ctany
再将y=x/2+45º代回,∫1/(1+sinx)dx=ctan(x/2+45º)
所以,∫sinx/(1+sinx) dx=x-ctan(x/2+45º)+c
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