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How do you find the antiderivative of $\cos \left( {{x^2}} \right)?$

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint: Find the integration of the given function in order to find its antiderivative. The given trigonometric expression is non elementary, so use the Maclaurin power expansion of the given function and then integrate that expansion to the antiderivative of the given function. Maclaurin expansion of cosine is given as
$\cos x = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n)!}}{x^{2n}}} $
Use this information to find the antiderivative of the given function.
Formula used:
Maclaurin series of cosine $\cos x = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n)!}}{x^{2n}}} $

Complete step by step solution:
To find the antiderivative of $\cos \left( {{x^2}} \right)$, let us understand first what is antiderivative of a function?
When we take an antiderivative also called the inverse derivative of a function f is a function then it gives a function F whose derivative is equal to the original function f.
That is in simple words antiderivative is the integration of a function.
Therefore to find the antiderivative of $\cos \left( {{x^2}} \right)$, we will find its integration.
Since the integral of $\cos \left( {{x^2}} \right)$ is non elementary, we cannot integrate it directly, so we will integrate it via power series.
For integration via power series, let us recall the Maclaurin series of cosine,
\cos x = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n)!}}{x^{2n}}} \\
\Rightarrow \cos \left( {{x^2}} \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{{( -
1)}^n}}}{{(2n)!}}{{\left( {{x^2}} \right)}^{2n}}} \\
\therefore \cos \left( {{x^2}} \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{{( -
1)}^n}}}{{(2n)!}}{x^{4n}}} \\
Taking the integration both sides, we will get
$\int {\cos \left( {{x^2}} \right)dx} = \int {\sum\limits_{n = 0}^\infty {\dfrac{{{{( -
1)}^n}}}{{(2n)!}}{x^{4n}}} } dx$
Here taking out the constant term,
\int {\cos \left( {{x^2}} \right)dx} = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n)!}}} \int
{{x^{4n}}} dx \\
= \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}}}{{(2n)!}}} \times \dfrac{{{x^{4n + 1}}}}{{(4n + 1)}} +
C \\
= \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}{x^{4n + 1}}}}{{(2n)!(4n + 1)}}} + C \\
Therefore $\sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^n}{x^{4n + 1}}}}{{(2n)!(4n + 1)}}} + C$ is the required antiderivative of $\cos \left( {{x^2}} \right)$

Note: There is a bit difference in antiderivative and integration, Integration is a function associates with the original function whereas antiderivative of $f(x)$ is just a function whose derivative is $f(x)$ This question can be solved with one more method in which we integrate $\cos \left( {{x^2}} \right)$ with help of the Fresnel integral. You will get a different expression at the end of the Fresnel integral process for this question but don’t worry both are equal and correct.