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

#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)#

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 #