Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 589: 50

$S = \mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 3} \right)\left( {2n + 1} \right)}}$ ${S_{34}} \approx 0.297953$

From Table 2: ${\tan ^{ - 1}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{2n + 1}} = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \le 1$. Substituting ${x^2}$ for $x$ in the series above gives ${\tan ^{ - 1}}\left( {{x^2}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {{x^2}} \right)}^{2n + 1}}}}{{2n + 1}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 2}}}}{{2n + 1}} = {x^2} - \dfrac{{{x^6}}}{3} + \dfrac{{{x^{10}}}}{5} - \dfrac{{{x^{14}}}}{7} + \cdot\cdot\cdot$ converges for $\left| {{x^2}} \right| \le 1$ or $\left| x \right| \le 1$. Taking the definite integral on the series gives $\mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^1 \dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 2}}}}{{2n + 1}}{\rm{d}}x$ $ = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 3}}}}{{\left( {4n + 3} \right)\left( {2n + 1} \right)}}} \right]_0^1 = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 3} \right)\left( {2n + 1} \right)}}$ Write $S = \mathop \smallint \limits_0^1 {\tan ^{ - 1}}\left( {{x^2}} \right){\rm{d}}x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {4n + 3} \right)\left( {2n + 1} \right)}}$. The alternating series above has positive terms: ${b_n} = \dfrac{1}{{\left( {4n + 3} \right)\left( {2n + 1} \right)}}$. By Eq. (2) in Section 11.4: $\left| {S - {S_n}} \right| \lt {b_{n + 1}}$ We choose $n$ so that $\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 4}}$ $\dfrac{1}{{\left( {4n + 7} \right)\left( {2n + 3} \right)}} \lt {10^{ - 4}}$ $\left( {4n + 7} \right)\left( {2n + 3} \right) \gt {10^4}$ Solving the inequality above we obtain $n \gt 33.73$. So we choose $N=34$ so that the series value is within an error of at most ${10^{ - 4}}$ to the definite integral. We compute the following: ${S_{34}} \approx 0.297953$, ${\ \ \ \ \ }$ $S \approx 0.297903$ It is verified that $\left| {S - {S_{34}}} \right| \approx 0.000051 \lt {10^{ - 4}}$.