What is the integral of $(\cos x{)}^2?$

Explanation:
We will use the cosine double-angle identity in order to rewrite ${\cos }^{2}x$. (Note that ${\cos }^{2}x=(\cos x{)}^2$, they are different ways of writing the same thing.)

$\cos (2x)=2{\cos }^{2}x-1$

This can be solved for ${\cos }^{2}x$ :

${\cos }^{2}x=\frac{\cos (2x)+1}{2}$
Thus,

$\int {\cos }^{2}xdx=\int \frac{\cos (2x)+1}{2}dx$

Split up the integral:

$=\frac{1}{2}\int \cos (2x)dx+\frac{1}{2}\int dx$

The second integral is the "perfect integral:" $\int dx=x+C$.

$=\frac{1}{2}\int \cos (2x)dx+\frac{1}{2}x$

The constant of integration will be added upon evaluating the remaining integral.
For the cosine integral, use substitution. Let $u=2x$, implying that $du=2dx$.
Multiply the integrand 2 and the exterior of the integral by $\frac{1}{2}$.

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

Substitute in $u$ and $du$ :

$=\frac{1}{4}\int \cos (u)du+\frac{1}{2}x$

Note that $\int \cos (u)du=\sin (u)+C$.

$=\frac{1}{4}\sin (u)+\frac{1}{2}x+C$

Since $u=2x$

$=\frac{1}{4}\sin (2x)+\frac{1}{2}x+C$

Note that this can be many different ways, $\mathrm{since}\sin (2x)=2\sin x\cos x$.