## Introduction to integral of sin x by x

In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.

Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute sin x/x integral by using different integration techniques.

## What is the integral of sin x/x?

The integral of sin x/x is an antiderivative of the sine function which is equal to Taylor’s series expansion of sine series. It is also known as the reverse derivative of the sine function, a trigonometric identity.

The sine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:

Sin = opposite side / hypotenuse

### Integral of sinx/x formula

The formula of the integral of sin contains the integral sign, coefficient of integration, and the function as sine. It is denoted by ∫(sin x/x)dx. In mathematical form, the integral of sin x/x is:

$∫\frac{\sin x}{x}dx=x-\frac{x^3}{3×3!}+\frac{x^5}{5×5!}-\frac{x^7}{7×7!}+ ...+ C$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral.

## How to calculate the integral of sin x/x?

The integral of sin x/x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate the integral of sinx/x by using:

- Taylor’s Series
- Integration by parts
- Laplace Transformation

## Integral of sinx/x by using Taylor’s Series

Taylor’s series is an infinite sum of terms that are expressed in terms of a function’s derivative. It can be used to calculate derivative of a function that is complex to solve. Since sin x/x is impossible to integrate by using formal integration. Therefore, we will use Taylor’s series to find the integral of sinx/x.

### Proof of integral of sin x by using Taylor’s Series

Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integral of sin x by x by using Taylor’s series. For this, we have to first assume the sine series that is,

$\sin x = x – \frac{x^3}{3!}+\frac{x^5}{5!}–\frac{x^7}{7!}+…$

We can use the above series in the integral of sin x to calculate the integral of sin x/x. Then,

$I = ∫\frac{\sin x}{x}dx$

Substituting the series of sinx, we get,

$I = ∫\frac{\left(x – \frac{x^3}{3!}+\frac{x^5}{5!}–\frac{x^7}{7!} +…\right)}{x}dx$

Now we can divide x in the denominator with each term of the sine series. The remaining terms are:

$I = ∫\left[1-\frac{x^2}{3!}+\frac{x^4}{5!}–\frac{x^6}{7!}+… \right]dx$

Now integrating the remaining terms, we get

$∫\frac{\sin x}{x}dx =x -\frac{x^3}{3×3!}+\frac{x^5}{5×5!}-\frac{x^7}{7×7!}+...+ C{2}$

Hence the above equation is the integration of sin x/x by using Taylor’s series. Use our Fourier series calculator to approximate any periodic function.

## Integral of sin x by using integration by parts

The method of integration by parts is a technique of solving integrals having a combination or product of two functions. This method is applicable to find the integral of complex functions. Therefore, we can compute the integral to sinx/x by using integration by parts.

### Proof of Integral of sin x by using integration by parts

Since the integral of sin x/x can be written as the product of two functions. Therefore, suppose that

$I = ∫\frac{\sin x)}{x}dx$

Using the method of integration calculator which is written as:

$∫{u.v}dx = u∫vdx - ∫[u’.∫vdx]dx$

Then,

$I =\left(-\frac{1}{x}\right)\cos x–∫x^{-2}.\cos xdx$

Integrating again,

$I=-\frac{\cos x}{x}–x^{-2}\sin x–∫\frac{2}{x^3}\sin x.dx$

We can substitute I in the above equation as:

$I=-\frac{\cos x}{x}–x^{-2}\sin x–I∫\frac{2}{x^2}dx$

Integrating the remaining term,

$I=-\frac{\cos x}{x}–x^{-2}\sin x + 2I(x^{-1})$

Taking 1/x common we get,

$I = \frac{1}{x}\left[-\cos x – \frac{\sin x}{x}+2I\right]$

More simplification,

$xI=\left[-\cos x–\frac{\sin x}{x} + 2I\right]$

$xI – 2I = - \cos x – \frac{\sin x}{x}{2}lt;/p>

$I(x – 2) = -\cos x –\frac{\sin x}{x}{2}lt;/p>

We can do more simplification as:

$I =\frac{-\cos x – \frac{\sin x}{x}}{(x – 2)}{2}lt;/p>

Hence the above is the verification of the integral of sin x/x by using integration by parts.

## Integral of sin x/x by using Laplace Transform

The Laplace transform is an integral transformation that is used to solve differential equations and integrals. Therefore, we can use integral transform to integrate sin x/x.

### Proof of integral of sin x/x by using Laplace Transform

To prove the integral of sin x/x by using Laplace Transform. We can use a definite integral to verify the integral of sin x/x. Assume the limit of the integral from 0 to infinity then,

$I=∫^∞_0\frac{\sin x}{x}dx$

Now, by using Laplace transformation, we can use the transformation of sin x such that:

$\mathscr{L}{\sin t} = \frac{1}{s^2} + 1$

Now, to get required integral,

$\mathscr{L}\frac{\sin t}{t}=∫^∞_0 \left[\frac{1}{s^2} + 1\right]ds$

The above equation is equal to,

$\mathscr{L}\frac{\sin t}{t}=∫^∞_0\left[\frac{1}{s^2} + 1\right]ds = \cot^{-1}s$

Now by using Laplace transform on the given integral,

$∫^∞_0 \frac{e^{-st}\sin t}{t} dt =∫^∞_0 \left[\frac{1}{s^2} + 1\right]ds = \cot^{-1}s$

Setting s = 0 we get,

$∫^∞_0 \frac{e^{-st}\sin t}{t} dt=\frac{π}{2}$

Hence we have verified the integral of sin x/x by using Laplace Transform. Since the above calculations are tricky and long-term, to avoid this try our Laplace transform calculator. It provides you a step-by-step solution within a few seconds.