Simplifying $(1+\cos2x)^2+(-2\sin2x)^2=1+2\cos2x+\cos^22x+4\sin^22x$ I'm supposed to find the arc length of this polar curve : $r=1+\cos2x$  where $0\leq x \leq \pi/4$.
I know that I have to use this formula $s=\displaystyle\int_{0}^{\pi/4} \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2} 
 dx$ My teacher always tells us to simplify the expression under the square root before taking the integral. But I'm stuck.
$(1+\cos2x)^2+(-2\sin2x)^2=1+2\cos2x+\cos^22x+4\sin^22x$
I have tried to use half-angle formulas and everything, but the expression does not get easier. Does someone have any tips?

For $r(x) = 1+\cos(2x)$, you have that your expression reads
$$ \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2} = \cos(x) \sqrt{10 - 6 \cos(2 x)}  = \cos(x) \sqrt{4 + 12 \sin(x)^2} $$
The integral now can be done by using $y=\sin(x)$ and $dy = \cos(x) dx$. You should obtain
$$
s = \sqrt{5}/2 + \tanh^{-1} \left( \sqrt{3/5} \right) /\sqrt{3}
$$ 
Edit: to simplify use the "double-angle formulae"
$(1+\cos2x)^2+(-2\sin2x)^2 = 1 +2 \cos 2x+ \cos^2 2x + 4 \sin^2 2x \\
= 2 + 2 \cos 2x+ 3 \sin^2 2x \\
= 4 \cos^2 x + 3 \sin^2 2x  \\
= 4 \cos^2 x + 12 \sin^2 x  \cos^2 x$.
This sequence of passages is not unique, other intuitions are possible.
