How do you integrate #int (-8x^3)/sqrt(9-x^2)dx# using trigonometric substitution?
1 Answer
Replace
Explanation:
Let
With these substitutions, we get
# = 8*27intcos^3 theta d theta#
# = 8*27int(1-sin^2 theta) cos theta d theta#
# = 8*27(sin theta-sin^3 theta / 3)+C#
With
# = 8*27(sqrt(9-x^2)/3 -((9-x^2)sqrt(9-x^2))/81)+C#
# = 8*27((27sqrt(9-x^2))/81 -((9-x^2)sqrt(9-x^2))/81)+C#
# = 8/3sqrt(9-x^2)(27 -(9-x^2))+C#
# = 8/3sqrt(9-x^2)(x^2+18)+C#