I'm working through an example from Stewart's book, and I can not work out why the lower bound of integration changes to a $0$. Also I am confused as to where the $8$ comes from?
I am assuming there is a rule being applied here that I am unaware of. I can't find any similar examples.
$$4\int_{-\pi/2}^{\pi/2}\cos^4\theta\ d\theta=8\int_0^{\pi/2}\cos^4\theta\ d\theta$$
On
$$\int_{-\pi/2}^{\pi/2} \cos^4 \theta d\theta = \int_{-\pi/2}^{0} \cos^4 \theta d\theta + \int_{0}^{\pi/2} \cos^4 \theta d\theta$$ Then let $t = -\theta$ in the first term on the RHS, then $dt = -d\theta$ and $\theta = -\pi/2 \implies t = \pi/2$ and $t =0 \implies \theta =0$, so a $u$-substitution gives that: $$ \int_{-\pi/2}^{0} \cos^4 \theta d\theta = \int_{\pi/2}^0 -\cos^4 (-t) dt = - \int_0^{\pi/2} - \cos^4 t dt = \int_0^{\pi/2} \cos^4 t dt.$$ Note we used that $\cos^4(-t) = \cos^4 t$.
On
The integral can be though of as the purple shaded area in the figure below.
Because the integrand is even (it is symmetric across the $y$-axis), the area on the left of the $y$-axis (the blue area) is equal to the area on the right side of the $y$-axis (the blue area). The area of the red region is the integral over the interval $[0,\pi/2]$.
More generally, we have the following:
Theorem: Suppose that $f : [-L,L] \to\mathbb{R}$ be an even, continuous function. Then $$ \int_{-L}^{L} f(x)\,\mathrm{d}x = 2 \int_{0}^{L} f(x)\,\mathrm{d}x. $$
Proof: By the linearity of integration, $$ \int_{-L}^{L} f(x) \,\mathrm{d}x = \int_{-L}^{0} f(x)\,\mathrm{d}x + \int_{0}^{L} f(x)\,\mathrm{d}x. \tag{1}$$ Making the change of variables $u = -x$, the first term becomes $$ \int_{-L}^{0} f(x)\,\mathrm{d}x = -\int_{L}^{0} f(-u)\,\mathrm{d}u = \int_{0}^{L} f(-u)\,\mathrm{d}u. $$ Since $f$ is even, $f(-u) = f(u)$ for any $u \in [-L,L]$. Therefore $$ \int_{0}^{L} f(-u)\,\mathrm{d}u = \int_{0}^{L} f(u)\,\mathrm{d}u. $$ Substituting this back into (1), we obtain \begin{align}\int_{-L}^{L} f(x) \,\mathrm{d}x &= \int_{-L}^{0} f(x)\,\mathrm{d}x + \int_{0}^{L} f(x)\,\mathrm{d}x \\ &= \int_{0}^{L} f(u)\,\mathrm{d}u + \int_{0}^{L} f(x)\,\mathrm{d}x \tag{these terms are equal}\\ &= 2\int_{0}^{L} f(x)\,\mathrm{d}x. \end{align} The last identity holds, as $u$ and $x$ are just dummy variables. After integration, the result is a constant which does not depend on either of these variables.
On
It is a polar curve looking somewhat like an egg/ellipse through origin with tangent there along y-axis having symmetry about x-axis.
The integral is the area of a (not mentioned) given polar curve $ r = 4\cos^2\theta.$
And for every such curve representing an even function
$$ \int_{-\pi/2} ^{\pi/2}\frac12 r^2 d \theta =2 \int_{0} ^{\pi/2}\frac12 r^2 d \theta =2 \int_{-\pi/2} ^{0}\frac12 r^2 d \theta. $$
Imagine trying to find the area of a circle. It suffices to find the area of a semicircle, then doubling it, right?
For any function that satisfies $$f(x)=f(-x),$$ we have $$\int_{-a}^af(x)\,dx=2\int_0^af(x)\,dx.$$ We call these functions even functions.