Show that $\int_0^1 (\sin{x}-4/9)^3dx<0$ without a calculator. In this age of electronic devices, sometimes I like to challenge myself to find clever ways to find solutions to math problems that seem to require a calculator, without a calculator. (Here is a favorite example.)
I recently came up with this:

Without a calculator, show that $$\int_0^1 \left(\sin{x}-\frac{4}{9}\right)^3dx<0$$

It's a pretty close shave: my computer says the LHS $\approx −0.0000050...$.
The exact value of the LHS is:
$$\frac{1}{729}\left(972(2+\cos{1})\sin^4{0.5}-243(2-\sin{2})+432(1-\cos{1})-64\right)$$
I tried to use Maclaurin series, but that seems to be futile.
I also tried to find some kind of useful symmetry of the graph of $y=\sin{x}-\frac{4}{9}$, but to no avail.
Any clever way to do this? Just curious.

Using Maclaurin series, we find that for all positive real $x$, the Maclaurin series of $\sin x$ alternates between being larger and smaller than $\sin x$. This is true because: 1) the Maclaurin series is always tangent to $\sin x$ as it shares the same first and second derivative (and so the error behaves almost like $\pm x^3$ with even smaller higher-order terms) (2) look at the behaviour of the function to infinity, as given by the sign of the highest-order term. For example, with $n = 3$ the limit to infinity of the Taylor series is $-\infty$, and so $\sin x > x - x^3/3!$. Choosing $x = \pi/4$ for example, we can see this is true given the first 3 decimal places, if we know $1/\sqrt{2} \approx 0.707$ and overestimate $\pi$ as $22/7$.
Given $f(x) = (\sin x - 4/9)^3$ and $T_n (x)$ being the Taylor series of $\sin x - 4/9$ to $n$ terms, $|f(x)^3 - (T_n (x))^3|$ $< f(x) - (T_n (x))$ when both $f(x), T_n (x)$ are small and if $T_n(x) > f(x)$. The right hand side of this inequality is bounded upwards by the Lagrange error bound of $\sin x$. This is given by $\frac{M}{(n + 1)!} |x - c|^{n+1}$. With $M = 1$, $x = 1$ and $c = 0$ (to find the maximum error for all $x$ between $0$ and $1$), we get that $\frac{1}{(n + 1)!} < 5 \cdot 10^{-6}$.
Now to bound this integral upwards, the negative area must be smaller and the positive area must be larger. In other words, we need to find an upper bound of $f(x)$, which gives candidates $x = 1, 5, 9 \cdots$ (and $f(x) < T_n(x)$ also holds for these values of $n$). $n = 5$ does not satisfy the inequality and so the next candidate is $n = 9$.
The value of this integral is $-5.0079 \cdot 10^{-6}$ which is a remarkably good approximation.
So we need to show that:
$$\int_0^1 \left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}-\frac{4}{9}\right)^{3} \ dx < 0$$
which would be extremely tedious to do completely by hand, but you only need a pocket calculator to show this.
