Chapter 8.6, Problem 29E

To determine

To find: The determine the power series to approximation of ${\displaystyle {\int }_0^{0.1}x\arctan \left(3x\right)dx}$ .

Answer to Problem 29E

  ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(3x\right)}^{2n+1}}{3\left(2n+1\right)}}dx$

Explanation of Solution

Given information: ${\displaystyle {\int }_0^{0.1}x\arctan \left(3x\right)dx}$

Calculation:

  $\begin{array}{l}{\displaystyle {\int }_0^{0.1}x\arctan \left(3x\right)dx}\\ ={\displaystyle {\int }_0^{0.1}x\left({\displaystyle \int \frac{3}{1+9x^2}dx}\right)dx}\\ ={\displaystyle {\int }_0^{0.1}\left(3x\right)}{\displaystyle \int {\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(3x\right)}^{2n}}dx}\\ ={\displaystyle {\int }_0^{0.1}\left(3x\right)}{\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(3x\right)}^{2n+1}}{3\left(2n+1\right)}}dx\\ ={\displaystyle {\int }_0^{0.1}{\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{3^{2n+1}\cdot x^{2n+1}}{\left(2n+1\right)}}dx}\\ ={\left[{\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\cdot \frac{3^{2n+1}}{2n+1}\cdot \frac{x^{2n+3}}{2n+3}}\right]}_0^{0.1}+c\end{array}$

Put, $x=0$ ,

Then, $c=0$

Now,

  $\begin{array}{l}{\displaystyle {\int }_0^{0.1}x\arctan \left(3x\right)}\\ ={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\cdot \frac{3^{2n+1}}{\left(2n+1\right)}\cdot \frac{{\left(0.1\right)}^{2n+3}}{\left(2n+3\right)}}\\ =\frac{3}{1.3}{\left(0.1\right)}^3-\frac{3^3}{3.5}{\left(0.1\right)}^5+\mathrm{........}\left[\begin{array}{l}\text{avoiding the terms correct }\\ \text{up to }6\text{ decimal place}\end{array}\right]\\ =0.001-0.0000018\\ =0.000998\end{array}$

Hence, ${\displaystyle {\int }_0^{0.1}x\arctan }=0.000998$