Question

How do you use a Power Series to estimate the integral $\int\limits_0^{0.01} {\cos \left( {{x^2}} \right)dx}$ ?

Answer 1

Hint : In order to find the integral of the given function with the given limits we need to use the power series. According to Maclaurin’s theorem for the cosine and sine function for the angle $x$ (in radians) we know that:
$\cos \left( x \right) = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ......$
$\sin \left( x \right) = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .....$

Complete step by step solution:
We are given the function $\int\limits_0^{0.01} {\cos \left( {{x^2}} \right)dx}$ .We need to use Power series to estimate the integral of the given function.
From, Maclaurin’s theorem for the cosine and sine function for the angle $x$ (in radians) we know that:
$\cos \left( x \right) = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ......$
$\sin \left( x \right) = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .....$
But here we are given with $\cos \left( {{x^2}} \right)$ So let’s change the values accordingly.
$\cos \left( x \right) = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ......$
So, we can write ${x^2}$ instead of $x$ and we get:
$\cos \left( {{x^2}} \right) = 1 - \dfrac{{{{\left( {{x^2}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {{x^2}} \right)}^4}}}{{4!}} - ......$
According to the Question, Put the above value and integrate it with the given limits, and we get:
$\int\limits_0^{0.01} {\cos \left( {{x^2}} \right)dx} = \int\limits_0^{0.01} {\left( {1 - \dfrac{{{{\left( {{x^2}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {{x^2}} \right)}^4}}}{{4!}} - ......} \right)} dx$
Separating the terms and integrating it, we get:

$$\int\limits_0^{0.01} {\cos \left( {{x^2}} \right)dx} = \int\limits_0^{0.01} {\left( {1 - \dfrac{{{{\left( {{x^2}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {{x^2}} \right)}^4}}}{{4!}} - ......} \right)} dx \\\ = \int\limits_0^{0.01} {1dx} - \int\limits_0^{0.01} {\dfrac{{{{\left( {{x^2}} \right)}^2}}}{{2!}}dx} + \int\limits_0^{0.01} {\dfrac{{{{\left( {{x^2}} \right)}^4}}}{{4!}}dx} - ..... \\\ = \int\limits_0^{0.01} {dx} - \int\limits_0^{0.01} {\dfrac{{\left( {{x^4}} \right)}}{{2!}}dx} + \int\limits_0^{0.01} {\dfrac{{\left( {{x^8}} \right)}}{{4!}}dx} - ..... \\\ = \left[ x \right] _0^{0.01} - \left[ {\dfrac{{{x^5}}}{{5.2!}}} \right] _0^{0.01} + \left[ {\dfrac{{{x^9}}}{{9.4!}}} \right] _0^{0.01} - ..... \\\ = \left[ {0.01 - 0} \right] - \left[ {\dfrac{{{{0.01}^5}}}{{5.2!}} - \dfrac{{{0^5}}}{{5.2!}}} \right] + \left[ {\dfrac{{{{0.01}^9}}}{{9.4!}} - \dfrac{{{0^9}}}{{9.4!}}} \right] - ..... \\\ \approx 0.01 \;$$

From the Second term onwards, terms are too small, so that terms are negligible.
Therefore, using the Power series for cosine to estimate the integral function $\int\limits_0^{0.01} {\cos \left( {{x^2}} \right)dx}$ we get the value approx $\approx 0.01$ .
So, the correct answer is “ $\approx 0.01$ ”.

Note : If there was sine instead of cosine then use the formula of sine that is $\sin \left( x \right) = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .....$ and rest would be same.
Many other properties for cosine and sine functions can be derived from these expansions: Like:
$\sin \left( { - x} \right) = - \sin \left( x \right) \\\ \cos \left( { - x} \right) = \cos \left( x \right) \\\ \dfrac{{d\cos x}}{{dx}} = - \sin x \;$