$\int {\cos }^4\left(x\right) \mathit{\ⅆ}x$

$\= \int {\left(\frac{1\+\cos \left(2 x\right)}{2}\right)}^2 \mathit{\ⅆ}x$

$\= \frac{1}{4}\int \left(1\+2 \cos \left(2 x\right)\+{\cos }^2\left(2 x\right)\right) \mathrm{dx}$

$\= \frac{1}{4}\int 1 \mathrm{dx}\+\frac{1}{4}\int 2 \cos \left(2 x\right) \mathrm{dx}\+\frac{1}{4}\int {\cos }^2\left(2 x\right) \mathrm{dx}$

$\= \frac{x}{4}\+\frac{1}{4} \sin \left(2 x\right)\+\frac{1}{4}\int \left(\frac{1\+\cos \left(2\cdot 2 x\right)}{2}\right) \mathrm{dx}$

$\= \frac{x}{4}\+\frac{1}{4} \sin \left(2 x\right)\+\frac{1}{8}\int 1 \mathrm{dx}\+\frac{1}{8}\int \cos \left(4 x\right) \mathrm{dx}$

$\= \frac{x}{4}\+\frac{1}{4} \sin \left(2 x\right)\+\frac{x}{8}\+\frac{1}{32} \sin \left(4 x\right)$

$\= \frac{3}{8} x\+\frac{1}{4} \sin \left(2 x\right)\+\frac{1}{32} \sin \left(4 x\right)$

Table 6.2.2(a)   Solution based on the identity  ${\cos }^2\left(x\right)\=\left(1\+\cos \left(2 x\right)\right)\/2$

$\int {\cos }^4\left(x\right) \mathit{\ⅆ}x$

$\=\frac{1}{4} {\cos }^3\left(x\right)\sin \left(x\right)\+\frac{3}{4}\int {\cos }^2\left(x\right) \mathrm{dx}$

$\=\frac{1}{4} {\cos }^3\left(x\right)\sin \left(x\right)\+\frac{3}{4}\left(\frac{1}{2}\cos \left(x\right)\sin \left(x\right)\+\frac{1}{2}\int 1 \mathrm{dx}\right)$

$\=\frac{1}{4} {\cos }^3\left(x\right)\sin \left(x\right)\+\frac{3}{8} \cos \left(x\right)\sin \left(x\right)\+\frac{3}{8} x$

Evaluate the given integral

$\int {\cos }^4\left(x\right) \mathit{\ⅆ}x$ = $\frac{1}{4}{\cos \left(x\right)}^3\sin \left(x\right)\+\frac{3}{8}\cos \left(x\right)\sin \left(x\right)\+\frac{3}{8}x$

$\begin{array}{c}\int {\cos \left(x\right)}^4\ⅆx\hfill \\ & \text{=}& \frac{\int {\left(1+\cos \left(2x\right)\right)}^2\ⅆx}{4}\hfill & \left[\mathrm{rewrite}\,{\cos \left(x\right)}^2=\frac{1}{2}+\frac{\cos \left(2x\right)}{2}\right]\\ & \text{=}& \frac{\int {\cos \left(2x\right)}^2\ⅆx}{4}+\frac{\int \cos \left(2x\right)\ⅆx}{2}+\frac{\int 1\ⅆx}{4}\hfill & \left[\mathrm{rewrite}\,{\left(1+\cos \left(2x\right)\right)}^2={\cos \left(2x\right)}^2+2\cos \left(2x\right)+1\right]\\ & \text{=}& \frac{\int {\cos \left(2x\right)}^2\ⅆx}{4}+\frac{\int \cos \left(2x\right)\ⅆx}{2}+\frac{x}{4}\hfill & \left[\mathrm{constant}\right]\\ & \text{=}& \frac{\int {\cos \left(u\right)}^2\ⅆu}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}+\frac{x}{4}\hfill & \left[\mathrm{change}\,u=2x\,u\right]\\ & \text{=}& \frac{\int \cos \left(2u\right)\ⅆu}{16}+\frac{\int \frac{1}{2}\ⅆu}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}+\frac{x}{4}\hfill & \left[\mathrm{rewrite}\,{\cos \left(u\right)}^2=\frac{\cos \left(2u\right)}{2}+\frac{1}{2}\right]\\ & \text{=}& \frac{\int \cos \left(2u\right)\ⅆu}{16}+\frac{u}{16}+\frac{\int \cos \left(2x\right)\ⅆx}{2}+\frac{x}{4}\hfill & \left[\mathrm{constant}\right]\\ & \text{=}& \frac{\int \cos \left(2u\right)\ⅆu}{16}+\frac{3x}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}\hfill & \left[\mathrm{revert}\right]\\ & \text{=}& \frac{\int \cos \left(\mathrm{u1}\right)\ⅆ\mathrm{u1}}{32}+\frac{3x}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}\hfill & \left[\mathrm{change}\,\mathrm{u1}=2u\,\mathrm{u1}\right]\\ & \text{=}& \frac{\sin \left(\mathrm{u1}\right)}{32}+\frac{3x}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}\hfill & \left[\cos \right]\\ & \text{=}& \frac{\sin \left(2u\right)}{32}+\frac{3x}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}\hfill & \left[\mathrm{revert}\right]\\ & \text{=}& \frac{\sin \left(4x\right)}{32}+\frac{3x}{8}+\frac{\int \cos \left(2x\right)\ⅆx}{2}\hfill & \left[\mathrm{revert}\right]\\ & \text{=}& \frac{\sin \left(4x\right)}{32}+\frac{3x}{8}+\frac{\int \cos \left(u\right)\ⅆu}{4}\hfill & \left[\mathrm{change}\,u=2x\,u\right]\\ & \text{=}& \frac{\sin \left(4x\right)}{32}+\frac{3x}{8}+\frac{\sin \left(u\right)}{4}\hfill & \left[\cos \right]\\ & \text{=}& \frac{\sin \left(4x\right)}{32}+\frac{3x}{8}+\frac{\sin \left(2x\right)}{4}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.2.2(b)   Annotated stepwise evaluation of $\int {\cos }^4\left(x\right) \mathrm{dx}$ via the Integration Methods tutor

Assign the unevaluated integral to a name

$q_1≔\int {\cos \left(x\right)}^4\ⅆx$

$\∫{\cos \left(x\right)}^4\ⅆx$

Apply the trig identity ${\cos }^2\left(x\right)\=\left(1\+\cos \left(2 x\right)\right)\/2$

$q_2≔\mathrm{simplify}\left(\mathrm{algsubs}\left({\cos }^2\left(x\right)\=\left(1\+\cos \left(2 x\right)\right)\/2\,q_1\right)\right)$

$\frac{1}{4}\∫\left({\cos \left(2x\right)}^2\+2\cos \left(2x\right)\+1\right)\ⅆx$

Split the integral into three separate integrals

$q_3≔\mathrm{expand}\left(q_2\,\cos \left(2 x\right)\right)$

$\frac{1}{4}\∫{\cos \left(2x\right)}^2\ⅆx\+\frac{1}{2}\∫\cos \left(2x\right)\ⅆx\+\frac{1}{4}\∫1\ⅆx$

Apply the trig identity ${\cos }^2\left(2 x\right)\=\left(1\+\cos \left(4 x\right)\right)\/2$

$q_4≔\mathrm{algsubs}\left({\cos }^2\left(2 x\right)\=\left(1\+\cos \left(4 x\right)\right)\/2\,q_3\right)$

$\frac{1}{4}\∫\left(\frac{1}{2}\cos \left(4x\right)\+\frac{1}{2}\right)\ⅆx\+\frac{1}{2}\∫\cos \left(2x\right)\ⅆx\+\frac{1}{4}\∫1\ⅆx$

Split the first integral into two separate integrals, keeping $\cos \left(2 x\right)$ and $\cos \left(4 x\right)$ intact

$\mathrm{expand}\left(q_4\,\cos \left(2 x\right)\,\cos \left(4 x\right)\right)$

$\frac{1}{8}\∫\cos \left(4x\right)\ⅆx\+\frac{3}{8}\∫1\ⅆx\+\frac{1}{2}\∫\cos \left(2x\right)\ⅆx$

Table 6.2.2(c)   Command-based solution from first principles