How do you integrate #int (2-x)/sqrt(x^2-4)dx# using trigonometric substitution?
1 Answer
# int (2-x)/sqrt(x^2-4) \ dx = 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C #
Explanation:
We seek:
# I = int (2-x)/sqrt(x^2-4) \ dx #
I would not use trigonometric substitutions. Using linearity we can split this into two integrals:
# I = int (2)/sqrt(x^2-4) - (x)/sqrt(x^2-4) \ dx#
# \ \ = int (2)/sqrt(x^2-4) \ dx - int \ (x)/sqrt(x^2-4) \ dx#
Consider the first integral:
# I_1 = int (2)/sqrt(x^2-4) \ dx #
We can perform a substitution of the form:
# u = x/2 => (du)/dx = 1/2 #
Substituting into the integral, we get:
# I_1 = int (2)/sqrt((2u)^2-4) \ (2) du #
# \ \ \ = int (1)/sqrt(4(u^2-1)) \ du #
# \ \ \ = 2 int (1)/sqrt(u^2-1) \ du #
Which is a standard integral, so we can readily integrate and restore the substitution:
# I_1 = 2 ln|u+sqrt(u^2-1)| #
# \ \ \ = 2 ln |x/2 + sqrt(x^2/4-1) | #
# \ \ \ = 2 ln |x/2 + sqrt(1/4(x^2-4)) | #
# \ \ \ = 2 ln |x/2 + 1/2sqrt(x^2-4) | #
# \ \ \ = 2 ln |1/2(x + sqrt(x^2-4)) | #
# \ \ \ = 2 ln (1/2) + 2ln|x + sqrt(x^2-4)| #
Now, we consider the second integral:
# I_2 = int \ (x)/sqrt(x^2-4) \ dx#
We can perform a substitution of the form:
# u = x^2-4 => (du)/dx = 2x #
Substituting into the integral, we get:
# I_2 = int \ (1)/sqrt(u) \ (1/2) \ du#
# \ \ \ = 1/2 \ int \ (1)/sqrt(u) \ du#
Which is a standard integral, so we can readily integrate and restore the substitution:
# I_2 = (1/2 sqrt(u))/(1/2)#
# \ \ \ = sqrt(u)#
# \ \ \ = sqrt(x^2-4)#
And Combining these results, and introducing a constant of integration, we get:
# I = 2 ln (1/2) + 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C_1 #
# \ \ = 2ln|x + sqrt(x^2-4)| - sqrt(x^2-4) + C #