How do you integrate #int 1/sqrt(2-5x^2)# by trigonometric substitution?
1 Answer
Oct 18, 2017
# int \ 1/sqrt(2-5x^2) \ dx = 1/sqrt(5) \ arcsin (sqrt(5/2)x) + C #
Explanation:
We seek:
# I = int \ 1/sqrt(2-5x^2) \ dx #
Which, we can write as:
# I = int \ 1/sqrt(2(1-5/2x^2)) \ dx #
# \ \ = int \ 1/(sqrt(2)sqrt(1-(sqrt(5)/sqrt(2)x)^2)) \ dx #
# \ \ = 1/sqrt(2) \ int \ 1/(sqrt(1-(sqrt(5/2)x)^2)) \ dx #
We can now perform a substitution, Let:
# u = sqrt(5/2)x => (du)/dx = sqrt(5/2) #
# :. sqrt(2/5) (du)/dx = 1 #
Substituting into the integral, we get:
# I = 1/sqrt(2) \ int \ 1/(sqrt(1-u^2)) \ (sqrt(2/5)) \ du #
# \ \ = 1/sqrt(5) \ int \ 1/(sqrt(1-u^2)) \ du #
Which is a standard integral, so we have:
# I = 1/sqrt(5) \ arcsin u + C #
Restoring the substitution:
# I = 1/sqrt(5) \ arcsin (sqrt(5/2)x) + C #