Question

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

Answer 1

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

Explanation:

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`

Answer 2

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

Explanation:

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`