Integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$ 
How to evaluate this integral:
$$\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$$

I can take the $\cos(x)$ out of the square root by expanding $\sin(2x)$ so it becomes $$\int_{0}^{\frac{\pi}{2}} \cos(x) \sqrt{\cos^2(x) +4\sin^2(x)} \, dx$$
but still I have no clue how to solve it.

You can proceed as
$$\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{\cos^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1-\sin^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1+ 3\sin^2{(x)}}\quad dx$$
Let $\sin x=\frac{t}{\sqrt 3}\implies \cos x\ dx=\frac{1}{\sqrt 3}dt$
$$=\frac{1}{\sqrt3}\int_{0}^{\sqrt 3} \sqrt{1+ t^2} dt$$
