# Integral

• Oct 10th 2009, 02:18 AM
Dranalion
Integral
Hey,

I am looking to antiderive the following:

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

Any help is much appreciated,

Dranalion
• Oct 10th 2009, 03:42 AM
galactus
$\displaystyle \int (3-sin(3x))cos(3x)dx$

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

Now, it should be easier.

$\displaystyle \frac{1}{3}\int (3-u)du$
• Oct 10th 2009, 03:50 AM
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.
• Oct 10th 2009, 04:17 AM
mr fantastic
Quote:

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.

• Oct 10th 2009, 04:19 AM
Defunkt
Quote:

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