Final Answer
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(4n+1\right)}}{\left(4n+1\right)\left(2n\right)!}+C_0$
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Step-by-step Solution
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1
Rewrite the function $\cos\left(x^2\right)$ as it's representation in Maclaurin series expansion
$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\left(x^2\right)^{2n}dx$
2
Simplify $\left(x^2\right)^{2n}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $2n$
$\int\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}x^{4n}dx$
3
We can rewrite the power series as the following
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\int x^{4n}dx$
4
Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $4n$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^n}{\left(2n\right)!}\frac{x^{\left(4n+1\right)}}{4n+1}$
5
Multiplying fractions $\frac{{\left(-1\right)}^n}{\left(2n\right)!} \times \frac{x^{\left(4n+1\right)}}{4n+1}$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(4n+1\right)}}{\left(4n+1\right)\left(2n\right)!}$
6
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(4n+1\right)}}{\left(4n+1\right)\left(2n\right)!}+C_0$
Final Answer
$\sum_{n=0}^{\infty } \frac{{\left(-1\right)}^nx^{\left(4n+1\right)}}{\left(4n+1\right)\left(2n\right)!}+C_0$