Hey, I am looking to antiderive the following: $\displaystyle \int (3-sin{3x})(cos{3x})dx$ Any help is much appreciated, Dranalion
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$\displaystyle \int (3-sin(3x))cos(3x)dx$ Just make the sub $\displaystyle u=sin(3x), \;\ \frac{du}{3}=cos(3x)dx$ Now, it should be easier. $\displaystyle \frac{1}{3}\int (3-u)du$
Using this, does this give the answer: = sin(3x) - 1/6(sin^2(3x)) + c ? I am just checking to see if I followed it correctly.
Originally Posted by Dranalion Using this, does this give the answer: = sin(3x) - 1/6(sin^2(3x)) + c ? I am just checking to see if I followed it correctly. When you differentiate your answer do you get the integrand?
Originally Posted by Dranalion Using this, does this give the answer: = sin(3x) - 1/6(sin^2(3x)) + c ? I am just checking to see if I followed it correctly. $\displaystyle \frac{1}{3} \int(3-u)du = \frac{1}{3} (3u - \frac{u^2}{2}) + C = u - \frac{u^2}{6} + C = sin(3x) - \frac{sin^2(3x)}{6} + C$ so yes
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