Question

Evaluate the integral `int \ sqrt(x-x^2)/x \ dx`?

Answer 1

`int sqrt(x-x^2)/x "d"x =xsqrt(1/x-1)-arctan(sqrt(1/x-1))+C`.

Explanation:

Note that the integrand requires `x>0`. For `x>0`, `x=sqrt(x^2)`. Then rewrite the integrand as,

`sqrt(x-x^2)/x=sqrt((x-x^2)/x^2)`.
`sqrt(x-x^2)/x=sqrt(1/x-1)`.

Then denote the required integral by `I`.

`I=int sqrt(1/x-1) "d"x`.

Make the substitution `u=sqrt(1/x-1)`. Then `"d"x = -2x^2sqrt(1/x-1)"d"u`. Then `I` is transformed to,

`I = -2 int x^2 (1/x-1) "d"u`.

From the definition of `u`, `1/x-1=u^2`. Then `x=1/(1+u^2)`, `x^2=1/(1+u^2)^2`. Then,

`I = -2 int u^2/(u^2+1)^2 "d"u`,
`I = - int u * (2u)/(u^2+1)^2 "d"u`.

Integration by parts states `int f'(u)g(u) "d"u = f(u)g(u) - int f(u)g'(u)"d"u`. Let `f'(u)= 2u/(u^2+1)^2`, `g(u)=u`.

A good general integration result to be aware of is, for `n != -1`,

`int p'(x) [p(x)]^(n) "d"x = ([p(x)]^(n+1))/(n+1)+C`.

This can be easily verified by differentiating the LHS.

Then as `f'(u)` is in this form we conclude `f(u) = -1/(u^2+1)`. Then, by parts,

`I = -(-1/(u^2+1)*u - int -1/(u^2+1)"d"u)`,
`I = u/(u^2+1) - int 1/(u^2+1) "d"u`,

Another general integration result to be aware of is `int 1/(u^2+1) "d"u = arctan(u) + C`. Then,

`I = u/(u^2+1) - arctan(u)+C`.

We know `u=sqrt(1/x-1)`. Then `u^2+1=1/x`. We conclude,

`I=xsqrt(1/x-1)-arctan(sqrt(1/x-1))+C`.

Answer 2

`int \ sqrt(x-x^2)/x \ dx = 1/2 arcsin(2x-1)+sqrt(x - x^2) + C`

Explanation:

We seek:

`I = int \ sqrt(x-x^2)/x \ dx`

We can start by completing the square of the numerator to get:

`x-x^2 = -(x^2-x)`
`" " = -{(x-1/2)^2-(1/2)^2}`
`" " = 1/4-(x-1/2)^2`
`" " = 1/4{1-(2x-1)^2}`

So we can write:

`I = int \ sqrt(1/4{1-(2x-1)^2})/x \ dx`
`\ \ = 1/2 \ int \ sqrt(1-(2x-1)^2)/x \ dx`

Then comparing the numerator with `1-sin^2A` we could try a trig substitution of the form:

`sintheta = 2x-1 => cos theta (d theta)/dx = 2`, and `x=1/2(sin theta+1)`

Substituting this into the integral and we get:

`I = 1/2 \ int \ sqrt(1-sin^2 theta)/(1/2(sin theta+1)) \ 1/2cos theta \ d theta`
`\ \ = 1/2 \ int \ cos^2 theta/(sin theta+1) \ d theta`
`\ \ = 1/2 \ int \ (1-sin^2 theta)/(1+sin theta) \ d theta`
`\ \ = 1/2 \ int \ ((1+sin theta)(1-sin theta))/(1+sin theta) \ d theta`
`\ \ = 1/2 \ int \ 1-sin theta \ d theta`
`\ \ = 1/2 \ {theta+cos theta} + C`

And from the substitution we have:

`sin^2theta = (2x-1)^2 => cos^2 theta = 1 - (2x-1)^2`
`:. cos^2 theta = 1 - 4x^2+4x-1`
`" " = 4x - 4x^2`

And so we can reverse the substitution to get:

`I = 1/2 \ {arcsin(2x-1)+sqrt(4(x - x^2))} + C`
`\ \ = 1/2 arcsin(2x-1)+sqrt(x - x^2) + C`