作业帮1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
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1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx
∫(1/x^2)cos(1/x)dx
=-∫cos(1/x)d(1/x)
=-sin(1/x)+C
2.
∵(arcsinx)'=xf(x)=(1-x^2)^ (-1/2)
∴f(x)=[x (1-x^2)^ 1/2] ^(-1)
1/f(x)=x(1-x^2) ^1/2
∫1/f(x)dx =∫x(1-x^2) ^1/2dx
=-1/2∫(1-x^2)^ 1/2 d(1-x^2)
= -1/3(1-x^2) ^(3/2) + C
3.∫（lnx/x）dx
=∫（lnx）dlnx
=1/2(lnx)^2+C
3、∫lnx/xdx=∫lnxd(lnx)=1/2(lnx)^2+C
1、∫1/x^2cos(1/x)=-∫cos(1/x)d(1/x)=sin(1/x)+c
(1)
∫ (1/x^2)*cos(1/x) dx
= -∫ dsin(1/x)
= -sin(1/x) + C
(2)
y= arcsinx
siny =x
cosyy' =1
∫xf(x)dx =arcsinx+C
xf(x) = 1/√(1-x^2)
1/f(x) = x√(1-x^2)
∫[1/f(x...
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(1)
∫ (1/x^2)*cos(1/x) dx
= -∫ dsin(1/x)
= -sin(1/x) + C
(2)
y= arcsinx
siny =x
cosyy' =1
∫xf(x)dx =arcsinx+C
xf(x) = 1/√(1-x^2)
1/f(x) = x√(1-x^2)
∫[1/f(x)] dx =∫ x√(1-x^2) dx
let
x= sina
dx = cosa da
∫[1/f(x)] dx
=∫ x√(1-x^2) dx
=∫ sina (cosa)cosa da
=- ∫ (cosa)^2 dcosa
= - (cosa)^3/3 + C'
=- (1/3)(1-x^2)^(3/2) + C'
(3)
∫lnx/x dx
= ∫lnxdlnx
=(lnx)^2/2 + C
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