∫sin^2(3x) dx
Follow Math Help Forum on Facebook and Google+
$\displaystyle \int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 6xdx=\frac{x}{2}-\frac{\sin 6x}{12}+C$
I dont understand: $\displaystyle \int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx$ more detail please
Originally Posted by gamboo I dont understand: $\displaystyle \int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx$ more detail please power reduction identity ... $\displaystyle \sin^2{u} = \frac{1 - \cos(2u)}{2}$ derived from the double angle identity for cosine ... $\displaystyle \cos(2u) = 1 - 2\sin^2{u} $
Thanks to both of you.
View Tag Cloud