Question

How do you integrate `((4-x^2)^(1/2)) / x^2`?

Answer 1

Keeping trigonometric substitution in mind, the numerator is of the form:

`sqrt(a^2 - x^2)`

which resembles `sqrt(1-sin^2x)`. Thus, let:

`x = asintheta` where `a = 2`
`=> x = 2sintheta`
`x^2 = a^2sin^2theta = 4sin^2theta`
`sqrt(4-x^2) = sqrt(2^2-2^2sin^2theta) = 2costheta`
`dx = 2costhetad theta`

With this substitution, we get:

`= int (cancel(2)costheta)/(cancel(4)sin^2theta) cancel(2)costhetad theta`

`= int cot^2thetad theta`

`= int csc^2theta - 1d theta`

since `1 + cot^2theta = csc^2theta`, just like how `1 + tan^2theta = sec^2theta`.

The derivative of `cottheta` is `-csc^2theta`, so with some negative-sign manipulation, we can get this into a more simplified form:

`= -int 1-csc^2thetad theta`

`= -(int d theta - intcsc^2thetad theta)`

`= -(int d theta + int-csc^2thetad theta)`

`= -int d theta - int-csc^2thetad theta`

`= -theta - cottheta`

With `x = 2sintheta` and `sqrt(4-x^2) = 2costheta`:

`=> theta = arcsin(x/2)`
`=> cottheta = costheta/sintheta = (sqrt(4-x^2)/cancel(2))(cancel(2)/x)`
`= sqrt(4-x^2)/x`

So the final answer is:

`= color(blue)(-arcsin(x/2) - sqrt(4-x^2)/x + C)`