Solving integrals without the Fundamental Theorem of Calculus Every time I see people attempt to solve or catalogue integrals, the approach ends up being to simplify and reduce the integrand using various techniques to a point where the integrand is simple enough to have the Fundamental Theorem of Calculus. The only exceptions, it seems to me, are improper integrals of analytic functions which can be solved using Cauchy's Integral Formula and extremely simple functions like $f(x)=3x^2$ which can be solved using summation techniques like Faulhaber's formula in the Riemann sum.
My intention with this question is to make a catalogue of integrals that can be obtained using either clever tricks with the Riemann sum definition or the intuitive property of area under a line. I will add a few of my own, but it would be amazing to see people solve integrals like $\int_0^{x}\frac{1}{t^2+1}dt=\tan^{-1}(x)$ without the FTC.
 A: Here I prove the formulas
\begin{equation}
\int_0^x\cos(t)dt=\sin(x)\tag{1}
\end{equation}
\begin{equation}
\int_0^x\sin(t)dt=1-\cos(x)\tag{2}
\end{equation}
without the Fundamental Theorem of Calculus. To begin, let us examine the below diagram of a general $n$-gon inscribed in a circle of radius $1$. The area of $n$-gon is $A$ and the area of the wedges is $B$.

The parametric equation in polar coordinates for the circle, assuming that the point at angle $\theta$ is the origin, is $r=2\sin(\alpha)$. Thus, letting $R$ be the region containing all of the area of the circle except for the wedge directly to the left of the angle $\theta$ we get that
\begin{align*}
\iint_Rdxdy=\pi-B\tag{3}
\end{align*}
Substituting into polar coordinates (which does not require the FTC) yields that
\begin{align*}
\iint_Rdxdy&=\int_{\frac{\pi-\theta}{2}}^{\pi}\int_0^{2\sin(\alpha)}rdrd\alpha\\
&=\frac{1}{2}\int_{\frac{\pi-\theta}{2}}^{\pi}\left(2\sin(\alpha)\right)^2d\alpha\\
&=2\int_{\frac{\pi-\theta}{2}}^{\pi}\sin^2(\alpha)d\alpha\\
\end{align*}
Here, the integral $\int_0^{2\sin(\alpha)} rdr$ is computed by simply examining the area of the triangle with base $2\sin(\alpha)$ and height $2\sin(\alpha)$. It is well known that the sum of angles in an $n$-gon is $\pi(n-2)$, and since all angles in a regular $n$-gon the value of $\theta$ is $\pi\frac{n-2}{n}$. Continuing thus from above, we get that
\begin{align*}
\iint_Rdxdy&=2\int_{\frac{\pi-\theta}{2}}^{\pi}\sin^2(\alpha)d\alpha\\
&=2\int_{\frac{\pi}{n}}^{\pi}\sin^2(\alpha)d\alpha\\
&=2\int_{0}^{\pi}\sin^2(\alpha)d\alpha-2\int_{0}^{\frac{\pi}{n}}\sin^2(\alpha)d\alpha\\
&=\pi-2\int_{0}^{\frac{\pi}{n}}\sin^2(\alpha)d\alpha\tag{4}\\
\end{align*}
Where $2\int_{0}^{\pi}\sin^2(\alpha)d\alpha=\pi$ is obtained from the fact that the value of $\iint_Rdxdy$ is $\pi$ when $n=\infty$. We now use the fact that the area of the regular $n$-gon ($A$)is $\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$ to get that
$$B=\frac{\pi-\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)}{n}=\frac{\pi}{n}-\frac{1}{2}\sin\left(\frac{2\pi}{n}\right)$$
substituting into (3) and then back into (4), we get that
\begin{align*}
\pi-\iint_Rdxdy&=B\\
&=\frac{\pi}{n}-\frac{1}{2}\sin\left(\frac{2\pi}{n}\right)\\
&=2\int_0^{\pi/n}\sin^2(t)dt
\end{align*}
Thus, if $x$ is in the form $\frac{\pi}{n}$, the equation
$$\int_0^{x}\sin^2(t)dt=\frac{x}{2}-\frac{1}{4}\sin\left(2x\right)$$
Since both the RHS and LHS are analytic, the Identity Theorem from complex analysis says that the equality must be true for all values of $x$ since $\frac{\pi}{n}$ has an accumulation point. Using the double angle formula $\cos(2x)=1-2\sin^2(x)$ we can turn this into our familiar formula for $\int_0^{x}\sin(t)dt$, and then examining the integral $\int_0^{\frac{\pi}{2}-x}\sin(t)dt$ yields the formula for $\int_0^{x}\cos(t)dt$.
