Question

How do you evaluate `int sqrt(4 - x^2)/x^2`?

Answer 1

The integral equals `-sqrt(4 -x^2)/x - arcsin(x/2) + C`

Explanation:

Use trigonometric substitution. We have an integral with a square root of the form `sqrt(a^2 - x^2)`. Therefore, use the substitution `x = asintheta`.

In our case, `a = 2`. Our substitution will therefore be `x = 2sintheta`. Then `dx- 2costhetad theta`.

`=>intsqrt(4 - (2sintheta)^2)/(2sintheta)^2 * 2costheta d theta`

`=>intsqrt(4 - 4sin^2theta)/(4sin^2theta) * 2costheta d theta`

`=>intsqrt(4(1 - sin^2theta))/(4sin^2theta) * 2costheta d theta`

Use `sin^2x + cos^2x = 1`:

`=>intsqrt(4cos^2theta)/(4sin^2theta) * 2costheta d theta`

`=>int(2costheta)/(4sin^2theta) * 2costheta d theta`

`=>int (4cos^2theta)/(4sin^2theta) d theta`

`=>int cos^2theta/sin^2theta d theta`

Use `cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx`:

`=>int cot^2theta d theta`

Use `cot^2x + 1 = csc^2x`:

`=>intcsc^2theta - 1 d theta`

`=>intcsc^2theta d theta- int 1 d theta`

These are both trivial integrals.

`=>-cottheta - theta + C`

We know from our original substitution that `x/2 = sintheta`. We can deduce that `theta = arcsin(x/2)`. Therefore, the side adjacent `theta` is `sqrt(4 - x^2)` in measure. Thus, `cottheta = sqrt(4 - x^2)/x`.

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

Hopefully this helps!