Evaluating $\int \cos^4(x)\operatorname d\!x$

$$\cos^4(x) = \left(\dfrac{1+\cos(2x)}2 \right)^2 = \dfrac{1 + \cos^2(2x) + 2\cos(2x)}4 = \dfrac{1 + \dfrac{1+\cos(4x)}2 + 2\cos(2x)}4$$ which gives us $$\cos^4(x) = \dfrac{3 + 4 \cos(2x) + \cos(4x)}8$$ Now you should be able to integrate this off.

Using the reduction formulae,

$$\int\cos^nxdx=\frac{\cos^{n-1}x\sin x}n+\frac{n-1}n \int\cos^{n-2}xdx$$

Putting $n=2,$ $$\int\cos^2xdx=\frac{\cos x\sin x}2+\frac12 \int dx=\frac{\cos x\sin x}2+\frac12 x+C$$

Putting $n=4,$ $$\int\cos^4xdx=\frac{\cos^3x\sin x}4+\frac34 \int\cos^2xdx$$