1. ## easy integration question

sorry i am having a major brain fade..
how do i integrate
integrate (6sin(x))^2-3(cos(x))^2

2. Originally Posted by calc626
sorry i am having a major brain fade..
how do i integrate
integrate (6sin(x))^2-3(cos(x))^2
$\sin^2{x} = \frac{1 - \cos(2x)}{2}$

$\cos^2{x} = \frac{1 + \cos(2x)}{2}$

3. $\int 6\sin^2(x)-3\cos^2(x) ~dx = \int \left(6\left(\frac{1-\cos(2x)}{2}\right)-3\left(\frac{1+\cos(2x)}{2}\right)\right) ~dx$

Does this help?