# Express the definite integral as an infinite series and find its value to within

Express the definite integral as an infinite series and find its value to within an error of at most
${10}^{-4}$
${\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx$
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i1ziZ
Term-by-Term Differentiation and Integration
$F\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{\left(x-c\right)}^{n}$
has radius of convergence $R>0$. Then F is differentiable on $\left(c-R,c+R\right)$. Furtehermore, we can integrate and differentiate term by term. For $x\in \left(c-R,c+R\right)$
${F}^{\prime }\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}n{a}_{n}{\left(x-c\right)}^{n-1}$
$\int F\left(x\right)dx=A+\sum _{n=0}^{\mathrm{\infty }}\frac{{a}_{n}}{n+1}{\left(x-c\right)}^{n+1}$ (A any constant)
These series have the same radius of convergence R
Here we need to find the value of $F\left(x\right)={\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx$ we will use the expansion of the $\mathrm{cos}x$
From table 2, We have Maclaurin Series $f\left(x\right)=\mathrm{cos}x=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)!}=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots$ converges for for all x here x is replaced with ${x}^{2}$
${\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{\left({x}^{2}\right)}^{2n}}{\left(2n\right)!}$
${\mathrm{cos}x}^{2}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{{x}^{4n}}{\left(2n\right)!}dx$
${\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\in }ft{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}$
Now we need to find out F(1) error less than 0.0001
${\int }_{0}^{1}{\mathrm{cos}x}^{2}dx=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}\frac{1}{\left(4n+1\right)\left(2n\right)!}$
Above is alternating series with