1. ## Another integration

Int[(cos(3x))^2]

Can someone help me with this too? I'll return the thumbsup tomorrow. I'm really exhausted.

2. ## Re: Another integration

By the double angle formula for cosine: $\cos(2\theta) = 2\cos^2 \theta-1$. Add one to both sides, then divide both sides by two. So, $\cos^2 \theta = \dfrac{1+\cos(2\theta)}{2}$. Plug in $3x$ for $\theta$. Now, you have:

$\int \dfrac{1+\cos(6x)}{2}dx = \int\left( \dfrac{1}{2} + \dfrac{\cos(6x)}{2}\right)dx = \int \dfrac{1}{2}x^0 dx + \int \dfrac{\cos(6x)}{2}dx$

Now, you can apply the power rule on the first integral and use the substitution $u=6x$ for the second.