Question

Evaluate the indefinite integral:∫sqrt(10x−x^2)dx ?

Answer 1

`20/3x^(3/2)-1/2x^2+c`

Explanation:

`int" "sqrt(10x-x^2)" "dx`

Complete the square,

`int" "sqrt(25-(x-5)^2)" "dx`

Substitute `u=x-5`,

`int" "sqrt(25-u^2)" "du`

Substitute `u=5sin(v)` and `du=5cos(v)`

`int" "5cos(v) sqrt(25-25sin^2(v))" "dv`

Simplify,

`int" "(5cos(v)) (5cos(v))" "dv`

Refine,

`int" "25cos^2(v)" "dv`

Take out the constant,

`25int" "cos^2(v)" "dv`

Apply double angle formulae,

`25int" "(1+cos(2v))/2" "dv`

Take out the constant,

`25/2int" "1+cos(2v)" "dv`

Integrate,

`25/2(v+1/2sin(2v))"+c`

Substitute back `v=arcsin(u/5)` and `u=x-5`

`25/2(arcsin((x-5)/5)+cancel(1/2sin)(cancel(2arcsin)((x-5)/5)))"+c`

Simplify,

`25/2(arcsin((x-5)/5))+25/2((x-5)/5)+c`

Refine,

`25/2arcsin((x-5)/5)+5/2(x-5)+c`, where `c` is the constant of integration.

Tadaa :D

Answer 2

`=1/2(((x-5)sqrt(-5(x^2-10x+20))) + 25/2arcsin((x-5)/5) +c`

Explanation:

What is `int sqrt(10x - x^2) dx` ?

Note that the domain of the function being integrated is where the inner quadratic is positive, i.e. `x in [0, 10]`

This expression can be integrated using substitutions. Though a possible pathway for integration doesn't immediately present itself, if we compete the square, then a trigonometric substitution can be carried out:

`10x - x^2 = 25 - (x-5)^2`

Which, we notice, is in the classic trigonometric substitution form, i.e. the square of a number minus the square of a linear `x` function.

First, to get rid of the linear, we let `u = x-5`, which gives `du=dx`, so we can rewrite the above integral as:

`int sqrt(25-u^2) du`

Now for the second substitution, let `u = 5sintheta`, which changes the integral to:

`int sqrt(25 - 25sin^2theta) dx`
`= int abs(5costheta) dx` (we can ignore the absolute value brackets)

Of course, the `dx` isn't helping, so we differentiate the substitution equation to get: `du = 5costheta d theta`, so the integral becomes:

`25 int cos^2 theta d theta`

Now we can use a double angle formula to make integrating `cos^2 theta` easier:

`cos(2 theta) = 2cos^2theta -1`
`:. cos^2theta = 1/2(cos(2theta)+1)`
So the integral becomes:

`25/2 int cos(2theta) + 1 d theta`
`=25/2(1/2sin(2 theta) + theta) + c`
`= 25/2(sinthetacostheta + theta) + c` (using a double-angle formula)

Now, `sintheta = u/5 = (x-5)/5`
Hence, `cos theta = sqrt(1-u^2/25) = sqrt((-x^2+10x-20)/25)`
And, `theta = arcsin(u/5) = arcsin((x-5)/5)`

`int sqrt(10x - x^2) dx`
`= 25/2(((x-5)sqrt(-5(x^2-20x+20)))/25 + arcsin((x-5)/5)) +c`
`=1/2(((x-5)sqrt(-5(x^2-10x+20))) + 25/2arcsin((x-5)/5) +c`