How do you integrate sin^2(3x)cos^3(3x) dx?
Use the identity sin^2(y) + cos^2(y) = 1 to get the expression sin^2(3x) (1-sin^2(3x)) cos(3x) dx.Use the substitution u= sin(3x) by dividing the expression by the derivative, u’= 3cos(3x).The expression then becomes u^2 (1-u^2) (1/3) du. Now everything is in terms of u so we can expand and integrate.Expanding gives (1/3) u^2 - (1/3) u^4 du.The answer in terms of u is now (1/9) u^3 - (1/15) u^5 + C. Don’t forget the +C!!!Finally, substitute back into x to get (1/9) sin^3(3x) - (1/15) sin^5(3x) +C.
ZG
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