Evaluate the integral? : # int \ x^2/(sqrt(x^2-25))^5 \ dx#
2 Answers
Explanation:
Substitute:
to have:
use now the trigonometric identity:
If we restrict ourselves to the interval
To undo the substitution note that:
Then:
Now consider
and undoing the substitution:
so the expression is the same for
# 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
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
# 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
# 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 #