Integrating trig function I'm stuck at this problem:
$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$
I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2}  $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead end. Any tips?

Try a substitution with the last expression that you have

$$\int\sqrt{\sin^4(x)+4\sin^2(x)\cos^2(x)}dx\quad u=\cos(x)$$.

This leads us to the following integral where we do another substitution:

$$\int\sqrt{3u^2+1}du \quad\!\! \text{substitute}\quad\!\! u=\sqrt{\dfrac{a}{b}}\tan (v) \quad\!\!\text{where}\quad\!\! a=1 \quad\!\!\text{and} \quad\!\!b=3$$

Then, the integral becomes the following:

$$\int\dfrac{\sqrt{\tan^2(v)+1}+\sec^2(v)}{\sqrt{3}}$$

and the rest is more or less is using trig identities.
