Question

How do you integrate `int sqrt(arcsinx/(1-x^2)` using substitution?

Answer 1

Not only does this integral needs a standard Trig substitution, but also an inverse trig identity not commonly used.

Explanation:

`int(sqrt((sin^-1)/(1-x^2))) dx`

`x=cosu rArrdx=-sinudu`
`cos^1x=u`

`sin^-1 +cos^-1=pi/2`

`sin^-1x=(pi/2)-cos^-1u`
`sin^-1x=pi/2 -u`

substituting into the integral

`int(sqrt((pi/2-u)/(1-cos^2u)))xx(-sinudu)`

`-int((pi/2-u)^(1/2)/sqrt(sin^2u))sinudu`

`-int((pi/2-u)^(1/2)/(sinu))sinudu`

`sinu` is cancelled from numerator and denominator

`-int(pi/2-u)^(1/2)du`

integrating by inspection

`=2/3(pi/2-u)^(3/2)+C`

substitute back for `u`

`=2/3(pi/2-cos^-1x)^(3/2)+C`

`=2/3(sin^-1x)^(3/2)+C`

Answer 2

`2/3(arcsinx)^(3/2)+C`

Explanation:

`intsqrt(arcsinx/(1-x^2))dx`

Note that the square root can be split up:

`=intsqrtarcsinx/sqrt(1-x^2)dx`

Furthermore, note that we have an `arcsin` function and its derivative present:

`=int(arcsinx)^(1/2)(1/sqrt(1-x^2))dx`

Using substitution, where `u=arcsinx` and `du=1/sqrt(1-x^2)dx`:

`=intu^(1/2)du`

Using the typical power rule for integration:

`=u^(3/2)/(3/2)+C=2/3u^(3/2)+C=2/3(arcsinx)^(3/2)+C`