Question

$\int {\cos }^{4}2xdx$

Answer 1

$$\begin{gathered} \int {\cos }^{4}2xdx \\ =\int {\left({\cos }^{2}2x\right)}^{2}dx \\ =\int {\left(\frac{1+\cos 4x}{2}\right)}^{2}dx\left[\therefore {\cos }^{2}x=\frac{1+\cos 2x}{2}\right] \\ =\frac{1}{4}\int {\left(1+\cos 4x\right)}^{2}dx \\ =\frac{1}{4}\int \left(1+{\cos }^{2}4x+2\cos 4x\right)dx \\ =\frac{1}{4}\int \left[1+\left(\frac{1+\cos 8x}{2}\right)+2\cos 4x\right]dx \\ =\frac{1}{4}\int \left(\frac{3}{2}+\frac{\cos 8x}{2}+2\cos 4x\right)dx \\ =\frac{1}{4}\left[\frac{3x}{2}+\frac{\sin 8x}{16}+\frac{2\sin 4x}{4}\right]+C \\ =\frac{3x}{8}+\frac{\sin 8x}{64}+\frac{\sin 4x}{8}+\mathrm{C} \end{gathered}$$