Hint. One may use the Taylor series expansion
$$
\log (1+u)=u-\frac{u^2}2+\frac{u^3}3-\frac{u^4}4+\frac{u^5}5+\cdots, \quad |u|<1,
$$ giving, for $0<a<1$,
$$
\int_0^a\log (1+x^4)\:dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^a\frac{x^{4n}}n\:dx=\sum_{n=1}^\infty(-1)^{n-1}\frac{a^{4n+1}}{n(4n+1)}.
$$ Then, for $0<a<1$, using the classic inequality (for alternating series)
$$
\left|\sum_{n=n_0}^\infty(-1)^{n-1}\frac{a^{4n+1}}{n(4n+1)}\right|\leq \frac{a^{4n_0+1}}{n_0(4n_0+1)}, \quad n_0\geq1,
$$ you get
$$
\left|\int_0^a\log (1+x^4)\:dx-\sum_{n=1}^{n_0}(-1)^{n-1}\frac{a^{4n}}{n(4n+1)}\right|\leq\frac{a^{4n_0+1}}{n_0(4n_0+1)}.
$$ By applying it to $a=0.2$ and $n_0=2$, we obtain
$$
\left|\int_0^{0.2}\log (1+x^4)\:dx-\sum_{n=1}^2(-1)^{n-1}\frac{(0.2)^{4n}}{n(4n+1)}\right|\leq\frac{(0.2)^9}{2\times9}\approx 0.00000003.
$$
In addition to changing the "upper bound" $t$ of integral,
you can make the integrand depends on $t$. This allows you
to get rid of the tail of the integrand for finite $t$.
For any $n \in \mathbb{Z}_{+}$, let $f_n : [0,\infty) \to \mathbb{R}$ be the function
$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, & x \le \sqrt{n}\\0, & x \ge \sqrt{n}\end{cases}$$
For any fixed $x \in [0,\infty)$, it is easy to see $f_n(x) \le f_{n+1}(x)$ whenever $x \ge \sqrt{n}$.
When $x < \sqrt{n}$, we can apply AM $\ge$ GM to $n$ copies of $1 - \frac{x^2}{n}$ and one copy of $1$ and get
$$1 - \frac{x^2}{n+1} = \frac{n}{n+1}\left(1 - \frac{x^2}{n}\right) + \frac{1}{n+1} \ge \left( 1 - \frac{x^2}{n}\right)^{n/n+1} \implies f_{n+1}(x) \ge f_n(x)$$
This means $f_1(x), f_2(x), \ldots$ is a sequence of pointwise non-decreasing,
non-negative functions. Since its pointwise limit equals to $e^{-x^2}$, we can use Monotone convergence theorm to convert the integral on $e^{-x^2}$ to a limit of single variable:
$$\begin{align}\frac{\sqrt{\pi}}{2}
= \int_0^\infty e^{-x^2} dx
&= \int_0^\infty \lim_{n\to\infty} f_n(x) dx
\stackrel{\rm MCT}{=}
\lim_{n\to\infty} \int_0^\infty f_n(x) dx\\
&= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx
= \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\
&= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}
\end{align}$$
Best Answer
Use the well-known Taylor series for the function $\sqrt{1+x}$ and substitute $x\to x^3$ to get $$\sqrt{1+x^3}=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}.$$ This series converges for $\lvert x\rvert <1$. Now integrating the series yields $$\int_0^{1/2}\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}\,\mathrm{d}x=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}.$$
We want to find the biggest $a_{n+1}$ which is smaller than $10^{-4}$. This is the case for $n=2$, so $$\lvert \sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}-\sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}\rvert \le \frac{1}{163840}<10^{-4}.$$
So the final answer is $$\int_0^{1/2}\sqrt{1+x^3}dx\approx \sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}= \frac{3639}{7168}$$ where the approximation error is at most $10^{-4}$. See here for numerical verification.