Question

Find the integral of $\dfrac{\sin x}{x}$?

Answer 1

Hint: Assume the integral of the given expression as I. Use the expansion formula of the sine function given by the Maclaurin series as $\sin x=x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+\dfrac{{{x}^{7}}}{7!}+....$ and divide both the sides by x to get the value of the expression $\dfrac{\sin x}{x}$. Now, integrate both the sides with respect to $dx$ and use the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ to integrate various terms of x. Add the constant of integration (C) at the end to complete the answer.

Complete step-by-step solution:
Here we have been provided with the expression $\dfrac{\sin x}{x}$ and we are asked to integrate it. Here we will use the expansion formula of the sine function to get the answer. Let us assume the integral as I, so we have,
$I=\int{\dfrac{\sin x}{x}dx}$
Now, using the expansion formula of the sine function given by the Maclaurin series as $\sin x=x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+\dfrac{{{x}^{7}}}{7!}+....$ we get the expression $\dfrac{\sin x}{x}$ as: -
$\Rightarrow \dfrac{\sin x}{x}=1+\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+\dfrac{{{x}^{6}}}{7!}+....$
Therefore, substituting the above expression inside the integral sign we get,
$\begin{align}
  & \Rightarrow I=\int{\dfrac{\sin x}{x}dx} \\
 & \Rightarrow I=\int{\left( 1+\dfrac{{{x}^{2}}}{3!}+\dfrac{{{x}^{4}}}{5!}+\dfrac{{{x}^{6}}}{7!}+.... \right)dx} \\
\end{align}$
Separating the terms we get,
\[\Rightarrow I=\int{1dx}+\int{\dfrac{{{x}^{2}}}{3!}dx}+\int{\dfrac{{{x}^{4}}}{5!}dx}+\int{\dfrac{{{x}^{6}}}{7!}dx}+.......\]
Here we can write the constant 1 as ${{x}^{0}}$, so using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ we get,
\[\begin{align}
  & \Rightarrow I=\int{{{x}^{0}}dx}+\int{\dfrac{{{x}^{2}}}{3!}dx}+\int{\dfrac{{{x}^{4}}}{5!}dx}+\int{\dfrac{{{x}^{6}}}{7!}dx}+....... \\
 & \Rightarrow I=\dfrac{{{x}^{0+1}}}{0+1}+\dfrac{1}{3!}\times \dfrac{{{x}^{2+1}}}{2+1}+\dfrac{1}{5!}\times \dfrac{{{x}^{4+1}}}{4+1}+\dfrac{1}{7!}\times \dfrac{{{x}^{6+1}}}{6+1}+..... \\
 & \Rightarrow I=x+\dfrac{{{x}^{3}}}{3\times 3!}+\dfrac{{{x}^{5}}}{5\times 5!}+\dfrac{{{x}^{7}}}{7\times 7!}+..... \\
\end{align}\]
Since we are finding the indefinite integral so we need to add a constant of integration (c) in the integral obtained, so we get,
\[\therefore I=C+x+\dfrac{{{x}^{3}}}{3\times 3!}+\dfrac{{{x}^{5}}}{5\times 5!}+\dfrac{{{x}^{7}}}{7\times 7!}+.....\]
Hence, the above relation is our answer.

Note: You must remember the expansion formula of the functions $\ln \left( 1+x \right),\sin x,\cos x,\tan x,{{e}^{x}}$ because they are used in the chapter limit and differentiation. You may think why we haven’t used the integration by parts method using the ILATE rule to get the answer. You can use that rule but remember that the integral will carry on up to infinite terms even in that case also.