Int[(cos(3x))^2]
Can someone help me with this too? I'll return the thumbsup tomorrow. I'm really exhausted.
By the double angle formula for cosine: $\displaystyle \cos(2\theta) = 2\cos^2 \theta-1$. Add one to both sides, then divide both sides by two. So, $\displaystyle \cos^2 \theta = \dfrac{1+\cos(2\theta)}{2}$. Plug in $\displaystyle 3x$ for $\displaystyle \theta$. Now, you have:
$\displaystyle \int \dfrac{1+\cos(6x)}{2}dx = \int\left( \dfrac{1}{2} + \dfrac{\cos(6x)}{2}\right)dx = \int \dfrac{1}{2}x^0 dx + \int \dfrac{\cos(6x)}{2}dx$
Now, you can apply the power rule on the first integral and use the substitution $\displaystyle u=6x$ for the second.