Find the following indefinite integral as power series and then compute the definite integral
$$\int_{0}^{1} \cos (x^{2}) dx$$ with the accuracy $10^{-4}$.
I would love to share my ideas regarding this problem, but unfortunately I do not have any, because this topic is kind of new to me and we have not solved similiar problems in our classes. So any help will be greatly appreciated and some day I can give your contribution back to the community as well :).
PS: I hope that the problem is still understandable, did my best to translate it from Finnish to English.
Best Answer
Well, it is a very well known result that:
$$\cos(x)=\sum_{\text{k}\ge0}\frac{(-1)^\text{k}x^{2\text{k}}}{\left(2\text{k}\right)!}\tag1$$
So:
$$\cos\left(x^2\right)=\sum_{\text{k}\ge0}\frac{(-1)^\text{k}x^{4\text{k}}}{\left(2\text{k}\right)!}\tag2$$
So, for the integral we get:
$$\mathcal{I}:=\int_0^1\cos\left(x^2\right)\space\text{d}x=\sum_{\text{k}\ge0}\frac{(-1)^\text{k}}{\left(2\text{k}\right)!}\int_0^1x^{4\text{k}}\space\text{d}x=\sum_{\text{k}\ge0}\frac{(-1)^\text{k}}{\left(2\text{k}\right)!}\cdot\left[\frac{x^{4\text{k}+1}}{4\text{k}+1}\right]_0^1=$$ $$\sum_{\text{k}\ge0}\frac{(-1)^\text{k}}{\left(2\text{k}\right)!}\cdot\left(\frac{1^{4\text{k}+1}}{4\text{k}+1}-\frac{0^{4\text{k}+1}}{4\text{k}+1}\right)=\sum_{\text{k}\ge0}\frac{(-1)^\text{k}}{\left(4\text{k}+1\right)\cdot\left(\left(2\text{k}\right)!\right)}\tag3$$
Now, we want to find:
$$\left|\mathcal{I}-\sum_{\text{k}=0}^\text{n}\frac{(-1)^\text{k}}{\left(4\text{k}+1\right)\cdot\left(\left(2\text{k}\right)!\right)}\right|\le10^{-4}\space\Longrightarrow\space\text{n}\ge3\tag4$$
So:
$$\mathcal{I}\approx\sum_{\text{k}=0}^3\frac{(-1)^\text{k}}{\left(4\text{k}+1\right)\cdot\left(\left(2\text{k}\right)!\right)}=\frac{25399}{28080}\tag5$$