Question

What is the integral of $\cos \left( {{t}^{2}} \right)$?

Answer 1

In order to find the integral of $\cos \left( {{t}^{2}} \right)$, firstly we have to expand this in the form of Maclurian series in order to find the function to be integrated easily. After expanding the terms, we will be truncating it to find the nearest integral to zero and then comparing it with wolfram alpha gives us the required solution.

Complete step-by-step answer:
Now let us have a brief regarding the Maclaurin series. It is a power series that allows one to calculate an approximation of a function $f\left( x \right)$ $f\left( x \right)$ $f\left( x \right)$ for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. In many practical applications, it is equivalent to the function it represents.
Now let us start finding the integral of $\cos \left( {{t}^{2}} \right)$.
Firstly, we will be expanding it in the Maclurian series.
$\cos \left( {{t}^{2}} \right)={{\left| \left[ \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{x}^{2n}}}{\left( 2n \right)!}} \right] \right|}_{x={{t}^{2}}}}$

& \Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{t}^{4n}}}{\left( 2n \right)!}} \\\ & =1-\dfrac{{{t}^{4}}}{2}+\dfrac{{{t}^{8}}}{24}-\dfrac{{{t}^{12}}}{720}+\dfrac{{{t}^{10}}}{40320}dt \\\ \end{aligned}$$ Now we will be truncating the series at the fifth non zero term so that we will be getting the integral near to $$t=0$$ $$\begin{aligned} & \cos \left( {{t}^{2}} \right)\approx 1-\dfrac{{{t}^{4}}}{2}+\dfrac{{{t}^{8}}}{24}-\dfrac{{{t}^{12}}}{720}+\dfrac{{{t}^{10}}}{40320}dt \\\ & \Rightarrow t-\dfrac{{{t}^{5}}}{10}+\dfrac{{{t}^{9}}}{216}-\dfrac{{{t}^{13}}}{9360}+\dfrac{{{t}^{11}}}{443520} \\\ \end{aligned}$$ For $$t$$ close to zero Now we will be comparing, in order to compare we will be using the wolfram alpha to get $$\begin{aligned} & \int _{-1}^{1}\cos \left( {{t}^{2}} \right)dt=1.80904848... \\\ & \int _{-1}^{1}1-\dfrac{{{t}^{4}}}{2}+\dfrac{{{t}^{8}}}{24}-\dfrac{{{t}^{12}}}{720}+\dfrac{{{t}^{10}}}{40320}dt=1.80905009 \\\ \end{aligned}$$ The only drawback that exists with using the wolfram alpha is that it only works with $$t=0$$. There is no elementary solution for this function. $$\therefore $$ The integral of $$\cos \left( {{t}^{2}} \right)$$ is $$1.8090$$. **Note:** The Maclaurin Series is a Taylor series centered about 0. The Taylor series can be centered around any number a a a and is written as follows: $$\sum \text{ }n\text{ }=\text{ }0\text{ }\infty \text{ }f\left( \text{ }n\text{ } \right)\left( \text{ }a\text{ } \right)\left( \text{ }x\text{ }-\text{ }a\text{ } \right)\text{ }n$$ and $$n\text{ }!\text{ }=\text{ }f\text{ }\left( \text{ }a\text{ } \right)\text{ }+\text{ }f\text{ }\prime \text{ }\left( \text{ }a\text{ } \right)\text{ }\left( \text{ }x\text{ }-\text{ }a\text{ } \right)\text{ }+\text{ }f\text{ }\prime \text{ }\prime \text{ }\left( \text{ }a\text{ } \right)\text{ }2\text{ }!$$.