Math Help - Integral

1. Integral

Hey,

I am looking to antiderive the following:

$\int (3-sin{3x})(cos{3x})dx$

Any help is much appreciated,

Dranalion

2. $\int (3-sin(3x))cos(3x)dx$

Just make the sub $u=sin(3x), \;\ \frac{du}{3}=cos(3x)dx$

Now, it should be easier.

$\frac{1}{3}\int (3-u)du$

3. 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.

4. 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.
$\frac{1}{3} \int(3-u)du = \frac{1}{3} (3u - \frac{u^2}{2}) + C = u - \frac{u^2}{6} + C = sin(3x) - \frac{sin^2(3x)}{6} + C$ so yes