E etobias New member Joined Jul 5, 2005 Messages 1 Jul 5, 2005 #1 Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle. [Integral] x^3 * [square root](9-x^2)dx; x = 3 sin [theta] Click to expand...
Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle. [Integral] x^3 * [square root](9-x^2)dx; x = 3 sin [theta] Click to expand...
G Gene Senior Member Joined Oct 8, 2003 Messages 1,904 Jul 5, 2005 #2 Are you sure the integral isn't from 0 to 3? x=3sin(T) dx=3cos(T)dT x^3=27sin^3(T) = 27(1-cos^2(T))sin(T) sqrt(9-x^2) = sqrt(9-9sin(T)^2) = sqre(9(1-sin^2(T))) = sqrt(9cos^2(T)) = +3cos(T) x^3*sqrt(9-x^2) = {27(1-cos^2(T))sin(T)}*{+3cos(T)} Can you take it from there?
Are you sure the integral isn't from 0 to 3? x=3sin(T) dx=3cos(T)dT x^3=27sin^3(T) = 27(1-cos^2(T))sin(T) sqrt(9-x^2) = sqrt(9-9sin(T)^2) = sqre(9(1-sin^2(T))) = sqrt(9cos^2(T)) = +3cos(T) x^3*sqrt(9-x^2) = {27(1-cos^2(T))sin(T)}*{+3cos(T)} Can you take it from there?
