How do you evaluate the integral #int x^-2arcsinx#?

Answer 1

#intx^-2arcsinxdx=lnabs(x/sqrt(1-x^2))-arcsinx/x+C#

Use integration by parts. Let:

#{(u=arcsinx,=>,du=1/sqrt(1-x^2)dx),(dv=x^-2dx,=>,v=-x^-1):}#

Then:

#intx^-2arcsinxdx=-arcsinx/x+int1/(xsqrt(1-x^2))dx#
Just working with the remaining integral, let #x=sintheta#. This implies that #sqrt(1-x^2)=costheta# and #dx=costhetad theta#. Then:
#int1/(xsqrt(1-x^2))dx=int1/(sinthetacostheta)costhetad theta=intcscthetad theta#
Which is a commonly known integral. Also note that if #sintheta=x#, this is represented in a right triangle where the side opposite #theta# is #x# and the hypotenuse is #1#. The leg adjacent to #theta#, then, is #sqrt(1-x^2)#. From this right triangle we can say that #csctheta=1/x# and #cottheta=sqrt(1-x^2)/x#, which will be relevant:
#int1/(xsqrt(1-x^2))dx=-lnabs(csctheta+cottheta)=-lnabs(1/x+sqrt(1-x^2)/x#
Combining the denominators and inverting the fraction by bringing the #-1# outside the natural log inside as a #-1# power, and putting this result into the original expression we found from integration by parts, we find a final answer of:
#intx^-2arcsinxdx=lnabs(x/sqrt(1-x^2))-arcsinx/x+C#
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Answer 2

#intx^-2arcsinxdx=lnabs(x/(1+sqrt(1-x^2)))-arcsinx/x+C#

Let #x=sintheta#. This implies that #dx=costhetad theta# and that #theta=arcsinx#. Then:
#I=intx^-2arcsinxdx=intcsc^2theta(theta)(costhetad theta)=intthetacotthetacscthetad theta#

Then perform integration by parts. Let:

#{(u=theta,=>,du=d theta),(dv=cotthetacscthetad theta,=>,v=-csctheta):}#

So:

#I=uv-intvdu=-thetacsctheta+intcscthetad theta#

Which is a common integral:

#I=-thetacsctheta-lnabs(csctheta+cottheta)=(-theta)/sintheta-lnabs((1+costheta)/sintheta)#
Rewriting the natural logarithm by bringing the #-1# into the integral as a #-1# power:
#I=(-theta)/sintheta+lnabs(sintheta/(1+sqrt(1-sin^2theta))#
And since #sintheta=x#:
#I=lnabs(x/(1+sqrt(1-x^2)))-arcsinx/x+C#
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Answer 3

To evaluate the integral ∫ x^(-2)arcsin(x) dx, use integration by parts. Let u = arcsin(x) and dv = x^(-2) dx. Then differentiate u to find du and integrate dv to find v. After finding du and v, apply the integration by parts formula:

∫ u dv = uv - ∫ v du

Substitute the values of u, dv, du, and v into the formula and perform the integration. Then simplify the resulting expression to obtain the final answer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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