Which appropriate trigonometric substitution is needed to evaluate: intdx/sqrt(x^6-x^8) ? Thank you so much in advance! Greetings
3 Answers
int \ 1/(sqrt(x^6-x^8)) \ dx = 1/4ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (sqrt(1-x^2))/(2x^2) + C
Explanation:
We seek:
I = int \ 1/(sqrt(x^6-x^8)) \ dx
\ \ = int \ 1/(sqrt(x^6(1-x^2))) \ dx
\ \ = int \ 1/(x^3sqrt(1-x^2)) \ dx
Consider a substitution:
u^2 = 1-x^2 => 2u(du)/dx = -2x \ \ \ , andx^2 = 1-u^2
Giving us:
I = int \ (-1/(x))/( x^3sqrt(1-x^2)) \ (-x) \ dx
\ \ = - \ int \ 1/( (x^2)^2sqrt(1-x^2)) \ (-x) \ dx
\ \ = - \ int \ 1/( (1-u^2)^2 \ u) \ u \ du
\ \ = - \ int \ 1/( (1-u^2)^2) \ du
\ \ = - \ int \ 1/( (u+1)^2(u-1)^2) \ du
And we can perform a partial fraction decomposition (omitted), on the integrand, giving us:
1/( (u+1)^2(u-1)^2) -= 1/4{1/(u+1)+1/(u+1)^2-1/(u-1)+1/(u-1)^2}
Thus we can write:
-4I = int 1/(u+1)+1/(u+1)^2-1/(u-1)+1/(u-1)^2 \ du
And we can integrate to get:
-4I = ln|u+1| -1/(u+1) -ln|u-1| -1/(u-1)
:. 4I = ln|u-1| - ln|u+1| + 1/(u+1) + 1/(u-1)
\ \ \ \ \ \ \ \ = ln|(u-1)/(u+1)| + ((u-1)+(u+1))/((u+1)(u-1))
\ \ \ \ \ \ \ \ = ln|(u-1)/(u+1)| + (2u)/(u^2-1)
Then we restore the substitution:
4I = ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| + (2sqrt(1-x^2))/(1-x^2-1)
\ \ \ = ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (2sqrt(1-x^2))/(x^2)
Finally:
I = 1/4ln|(sqrt(1-x^2)-1)/(sqrt(1-x^2)+1)| - (sqrt(1-x^2))/(2x^2)
Explanation:
Applying Trigonometric Substitution
Simplify
let
Apply Integration by Parts
From
Reverse the Trigonometric Substitution
Simplify
I hope this was helpful.
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