Question

Evaluate the integral? : `int \ x^2/(sqrt(x^2-25))^5 \ dx`

Answer 1

`int x^2/(sqrt(x^2-25))^5 dx = -1/75 x^3/(sqrt(x^2-25))^3 + C`

Explanation:

`int x^2/(sqrt(x^2-25))^5 dx = int x^2/(5sqrt((x/5)^2-1))^5 dx`

`int x^2/(sqrt(x^2-25))^5 dx = 1/5^5int x^2/(sqrt((x/5)^2-1))^5 dx`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25int (x/5)^2/(sqrt((x/5)^2-1))^5 d(x/5)`

Substitute:

`x/5 = sect`

`d(x/5) = sect tan t`

to have:

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int sec^2t/(sqrt(sec^2t-1))^5 sect tant dt`

use now the trigonometric identity:

`sec^2t -1 = tan^2t`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int sec^2t/(sqrt(tan^2t))^5 sect tant dt`

If we restrict ourselves to the interval `x in (5,+oo)` so that `t in (0,pi/2)` we have that `tan t > 0` so `sqrt(tan^2t) = tant`:

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int sec^2t/tan^5t sect tant dt`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int sec^3t/tan^4t dt`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int 1/cos^3t cos^4t/sin^4t dt`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int cost/sin^4t dt`

`int x^2/(sqrt(x^2-25))^5 dx = 1/25 int (d(sint))/sin^4t`

`int x^2/(sqrt(x^2-25))^5 dx = -1/75 1/sin^3t + C`

To undo the substitution note that:

`sint = sqrt (1-cos^2t) = sqrt(1-1/sec^2t) = sqrt(1-25/x^2) = sqrt(x^2-25)/x`

Then:

`int x^2/(sqrt(x^2-25))^5 dx = -1/75 x^3/(sqrt(x^2-25))^3 + C`

Now consider `x in (-oo,-5)` and substitute `t= -x`:

`int x^2/(sqrt(x^2-25))^5 dx = int (-t)^2/(sqrt((-t)^2-25))^5 d(-t)`

`int x^2/(sqrt(x^2-25))^5 dx = -int t^2/(sqrt(t^2-25))^5 dt`

`int x^2/(sqrt(x^2-25))^5 dx = 1/75 t^3/(sqrt(t^2-25))^3 + C`

and undoing the substitution:

`int x^2/(sqrt(x^2-25))^5 dx = -1/75 x^3/(sqrt(x^2-25))^3 + C`

so the expression is the same for `x in (-oo,5) uu (5,+oo)`

Answer 2

`int \ x^2/(sqrt(x^2-25))^5 \ dx = -1/75 \ x^3/( sqrt(x^2 -25) )^3 + C`

Explanation:

Compare the denominator to the trig identity:

`tan^2A -=sec^2A-1`

In an attempt to reduce the denominator to something simper.

Why this identity in particular? Well firstly it is a trig identity of the form `fn1^2 = fn2^2-1` so with an appropriate factor it will reduce to being dependant upon `fn1` alone, and secondly, `d/dxtanx=sec^2x` and `d/dxsec=secxtanx` so either derivative also appears in the identity, which should help with a substitution in the integral,

So we want to find:

`I = int \ x^2/(sqrt(x^2-25))^5 \ dx`
`\ \ = int \ x^2/(sqrt(25(x^2/25-1)))^5 \ dx`
`\ \ = int \ x^2/(5sqrt((x/5)^2-1))^5 \ dx`

So let us try a substitution `sec theta = x/5` such that

`x=5sec theta`, and `dx/(d theta) = 5sec theta tan theta`

Thus:

`I = int \ (5sec theta)^2/(5(sqrt(sec^2 theta-1)))^5 5sec theta tan theta \ d theta`

`\ \ = int \ (5^3sec^3 theta tan theta)/(5(sqrt(tan^2theta)))^5 \ d theta`

`\ \ = int \ (5^3sec^3 theta tan theta)/(5^5tan^5 theta) \ d theta`
`\ \ = 1/25 \ int \ (sec^3 theta )/(tan^4 theta) \ d theta`

`\ \ = 1/25 \ int \ (1/cos^3 theta )/(sin^4 theta / cos^4 theta) \ d theta`

`\ \ = 1/25 \ int \ (cos theta )/(sin^4 theta ) \ d theta`

Which we can integrate by observation as:

`d/(du) 1/sin^3u = -3cosu/sin^4u`

And so we conclude that:

`I = (1/25 \ 1/sin^3theta)/(-3) + C`
`\ \ = -1/75 \ 1/sin^3theta + C`

If we refer back to our earlier substitution, we note:

`x=5sec theta => sec theta = x/5`
`cos theta = 5/x`

And using the trig identity `cos^2A + sin^2A -= 1` this gives:

`sin^2 theta + (5/x)^2 = 1 => sin theta = sqrt(1 -25/x^2)`

Using this to reverse the earlier substitution we get:

`I = -1/75 \ 1/(sqrt(1 -25/x^2))^3 + C`
`\ \ = -1/75 \ 1/(sqrt(1 -25/x^2))^3 + C`
`\ \ = -1/75 \ 1/( sqrt(1/x^2(x^2 -25)) )^3 + C`
`\ \ = -1/75 \ 1/( 1/xsqrt(x^2 -25) )^3 + C`
`\ \ = -1/75 \ x^3/( sqrt(x^2 -25) )^3 + C`