Question

How do you use the properties of integrals to verify the inequality $\int{\dfrac{\sin \left( x \right)}{x}}$ from $\dfrac{\pi }{4}$ to $\dfrac{\pi }{2}$ is less than or equal to $\dfrac{\sqrt{2}}{2}$?

Answer 1

In order to find the solution of the given question that is to verify that the inequality $\int{\dfrac{\sin \left( x \right)}{x}}$ from $\dfrac{\pi }{4}$ to $\dfrac{\pi }{2}$ is less than or equal to $\dfrac{\sqrt{2}}{2}$ use Maclaurin series of sine and substitute in the given integral and divide it by $x$ then solve the integral and apply the given value of limits of the integral.

Complete step by step answer:
According to the question, we have show that the inequality $\int{\dfrac{\sin \left( x \right)}{x}}$ from $\dfrac{\pi }{4}$ to $\dfrac{\pi }{2}$ is less than or equal to $\dfrac{\sqrt{2}}{2}$, in other words we can say that:
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}$
We should start by noting that the integral of $\dfrac{\sin \left( x \right)}{x}$ is not defined by real functions. Therefore, we will have to use the maclaurin series.
The maclaurin series for sine is known to be as follows:
$\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...$
To determine the maclaurin series for $\dfrac{\sin \left( x \right)}{x}$, we must divide each term by $x$ from the above expression, we get:
$\Rightarrow \dfrac{\sin \left( x \right)}{x}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n \right)}}}{\left( 2n-1 \right)!}}=1-\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+...$
Now to determine the value of $\int{\dfrac{\sin \left( x \right)}{x}}dx$, we must integrate term by term (with respect to $x$) we get:
$\Rightarrow \int{\dfrac{\sin \left( x \right)}{x}dx=}\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n+1 \right)}}}{\left( 2n+1 \right)\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C$
As for the definite integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx$, we simply use the second fundamental theorem of calculus to evaluate it as follows:
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\left[ x-\dfrac{{{x}^{3}}}{3\left( 3! \right)}+\dfrac{{{x}^{5}}}{5\left( 5! \right)}+...+C \right]_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx=\dfrac{\pi }{2}-\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{5}}}{600}-\left( \dfrac{\pi }{4}-\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{3}}}{18}+\dfrac{{{\left( \dfrac{\pi }{4} \right)}^{5}}}{600} \right)$
After computing the sum of the first few terms, we get $s=0.61$.
The integral converges to this value, because the more terms you add the less the decimals change (the actual value of the integral computed by calculator gives $0.611786287085$ which is less than $\dfrac{\sqrt{2}}{2}=0.707$.
$\Rightarrow \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}$
The integral will satisfy the inequality as long as you use more than one term of the maclaurin series in the approximation.
Therefore, we can have proved that $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( x \right)}{x}}dx\le \dfrac{\sqrt{2}}{2}$.

Note:
Students can go wrong by applying the wrong formula of maclaurin series of sine that is some student use $\sin \left( x \right)=x+\dfrac{{{x}^{3}}}{3!}-\dfrac{{{x}^{5}}}{5!}+...$ which is completely wrong and leads to the wrong answer. It’s important to remember here that correct formula of maclaurin series of sine is $\sin \left( x \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( -1 \right)}^{\left( n-1 \right)}}{{x}^{\left( 2n-1 \right)}}}{\left( 2n-1 \right)!}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+...$.