## Taylor Series Uses #1: Complicated Integrals

1. Can you compute $\displaystyle\int^{1}_{0}\cos(t^{2})\,\mathrm{d}t$?

Well, first we consider the antiderivative for $\cos(t^{2})$…which doesn’t exist.

What do we do? Cry. No, what I mean is, use Taylor series!

2. We should recall the Taylor series for cosine is
(1)$\displaystyle \cos(x)=\sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}x^{2n}$
So we plug in $x=t^{2}$ to the series and we get
(2)$\displaystyle \cos(t^{2})=\sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}t^{4n}$
Now we plug (2) into our integral, we get
(3)\begin{aligned} \int^{1}_{0}\cos(t^{2})\,\mathrm{d}t&=\int^{1}_{0}\sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}t^{4n}\,\mathrm{d}t\\ &=\sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}\int^{1}_{0}t^{4n}\,\mathrm{d}t\end{aligned}
This is a more manageable calculation, but we end up with an infinite series
(4)\begin{aligned} \int^{1}_{0}\cos(t^{2})\,\mathrm{d}t&=\sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}\int^{1}_{0}t^{4n}\,\mathrm{d}t\\ &=\sum^{\infty}_{n=0}\frac{(1)^{n}}{(2n)!}\frac{1}{4n+1}\end{aligned}
How do we evaluate this?

3. We usually truncate it at some point. For example, we know each term decreases (the series converges absolutely). So if we stop at the fifth term, we are throwing away terms smaller than
(5)$\displaystyle\frac{1}{10!}\frac{1}{21}\approx 0.000000013122533$
So to about 7 digits, this gives us a good approximation
(6)$\displaystyle \int^{1}_{0}\cos(t^{2})\,\mathrm{d}t\approx 0.9045242$
where we truncate to 7 digits (i.e., we throw away everything after 7 digits, since it’d be corrected by higher order terms in the series).