How do you integrate #int (-8x^3)/sqrt(9-x^2)dx# using trigonometric substitution?

1 Answer
Jim H
Mar 18, 2016

Replace #x# with either #3sintheta# or with #3costheta#

Explanation:

#int (-8x^3)/sqrt(9-x^2)dx#

Let #x=3costheta#. This makes

#dx=-3sintheta d theta#, and

#x^3 = 27cos^3 theta#, and

#sqrt(9-x^2) = 3sin theta#

With these substitutions, we get

#int (-8x^3)/sqrt(9-x^2)dx = int (-8(27cos^3theta))/(3sin theta) (-3sintheta) d theta#

# = 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 #cos theta = x/3#, we get #sin theta = sqrt(9-x^2)/3#, so our integral becomes:

#int (-8x^3)/sqrt(9-x^2)dx = 8*27(sqrt(9-x^2)/3 -(sqrt(9-x^2)/3)^3 / 3)#

# = 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#

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