Chapter 10, Problem 63RE

(a)

To determine

The series ${\displaystyle \underset{0}{\overset{x}{\int }}\mathrm{Sin}{t}^2}dt$ as power series.

Answer to Problem 63RE

The series is given as, ${\displaystyle \underset{0}{\overset{x}{\int }}\sin t^{2}dt}=\frac{x^3}{3}-\frac{x^7}{7\left(3!\right)}+\frac{x^{11}}{11\left(5!\right)}-\mathrm{.....}+{\left(-1\right)}^n\frac{x^{4n+3}}{\left(4n+3\right)\left(2n+1\right)!}+\mathrm{....}$

Explanation of Solution

Given information:

The Maclaurin series can be used to express the non-elementary signals in terms of series,

  ${\displaystyle \underset{0}{\overset{x}{\int }}\mathrm{Sin}{t}^2}dt$

Formula used:

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\mathrm{......}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{....}$

Calculation:

Maclaurin series generated by function sin x is given by

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\mathrm{......}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{....}$

Replacing x by t² in the above series,

  $\begin{array}{l}\sin t^2=t^2-\frac{{\left(t^2\right)}^3}{3!}+\frac{{\left(t^2\right)}^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{{\left(t^2\right)}^{2n+1}}{\left(2n+1\right)!}+\mathrm{...}\\ \sin t^2=t^2-\frac{t^6}{3!}+\frac{t^{10}}{5!}-\mathrm{.....}+{\left(-1\right)}^n\frac{t^{4n+2}}{\left(2n+1\right)!}+\mathrm{...}\end{array}$

Now, integrating term by term, observing that the constant term is zero,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{x}{\int }}\sin t^{2}dt={\displaystyle \underset{0}{\overset{x}{\int }}\left[t^2-\frac{t^6}{3!}+\frac{t^{10}}{5!}-\mathrm{.....}+{\left(-1\right)}^n\frac{t^{4n+2}}{\left(2n+1\right)!}+\mathrm{....}\right]dt}}\\ =\frac{x^3}{3}-\frac{x^7}{7\left(3!\right)}+\frac{x^{11}}{11\left(5!\right)}-\mathrm{....}+{\left(-1\right)}^n\frac{x^{4n+3}}{\left(4n+3\right)\left(2n+1\right)!}+\mathrm{.....}\end{array}$

Therefore,

  ${\displaystyle \underset{0}{\overset{x}{\int }}\sin t^{2}dt}=\frac{x^3}{3}-\frac{x^7}{7\left(3!\right)}+\frac{x^{11}}{11\left(5!\right)}-\mathrm{.....}+{\left(-1\right)}^n\frac{x^{4n+3}}{\left(4n+3\right)\left(2n+1\right)!}+\mathrm{....}$

Conclusion:

The series is given as, ${\displaystyle \underset{0}{\overset{x}{\int }}\sin t^{2}dt}=\frac{x^3}{3}-\frac{x^7}{7\left(3!\right)}+\frac{x^{11}}{11\left(5!\right)}-\mathrm{.....}+{\left(-1\right)}^n\frac{x^{4n+3}}{\left(4n+3\right)\left(2n+1\right)!}+\mathrm{....}$

(b)

To determine

The number of terms in ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$

Answer to Problem 63RE

The degree is 7.

Explanation of Solution

Given information:

The given series is ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$

Formula used:

Power series is used.

Calculation:

  ${\displaystyle \underset{0}{\overset{1}{\int }}\sin x^{2}dx}=\frac{1}{3}-\frac{1}{7\left(3\right)!}+\frac{1}{11\left(5!\right)}-\mathrm{....}$

Since the third term is

  $\frac{1}{11\left(5!\right)}=\frac{1}{1320}<0.001$

It suffices to use the first two non-zero terms. That is, up to degree 7.

Conclusion:

The degree is 7.

(c)

To determine

The value of ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$

Answer to Problem 63RE

The value is ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$ =0.31026830

Explanation of Solution

Given information:

The given series is ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$

Formula used:

NINT method is used.

Calculation:

NINT $\left(\sin x^2,x,0,1\right)\approx 0.31026830$

Conclusion:

The value is ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$ =0.31026830

(d)

To determine

To analyze: The four terms that is present in series.

Answer to Problem 63RE

The value is within $1.5\times {10}^{-7}$

Explanation of Solution

Given information:

The series is ${\displaystyle \underset{0}{\overset{1}{\int }}\mathrm{Sin}{x}^2}dx$

Calculation:

Using first four terms of series,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\sin x^{2}dx=\frac{1}{3}-\frac{1}{7\left(3!\right)}+}\frac{1}{11\left(5!\right)}-\frac{1}{15\left(7!\right)}\\ =\frac{258019}{831600}\\ {\displaystyle \underset{0}{\overset{1}{\int }}\sin x^{2}dx=0.31026816}\end{array}$

Therefore,

This is within $1.5\times {10}^{-7}$ of the answer in (c)

Conclusion:

The value is within $1.5\times {10}^{-7}$