sorry i am having a major brain fade.. how do i integrate integrate (6sin(x))^2-3(cos(x))^2
Last edited by mr fantastic; Dec 8th 2009 at 11:52 AM. Reason: rt --> de
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Originally Posted by calc626 sorry i am having a major brain fade.. how do i integrate integrate (6sin(x))^2-3(cos(x))^2 $\displaystyle \sin^2{x} = \frac{1 - \cos(2x)}{2}$ $\displaystyle \cos^2{x} = \frac{1 + \cos(2x)}{2}$
Last edited by mr fantastic; Dec 8th 2009 at 11:53 AM.
$\displaystyle \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?
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