## Introduction to integral of cosh 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 the coshx integral by using different integration techniques.

## What is the integral of cosh x?

The integral of cosh x is an antiderivative of the coshx function which is equal to sinh x. It is also known as the reverse derivative of the cosine function, a hyperbolic function. By definition, the hyperbolic function cos is a combination of two exponential functions e^x and e^-x. Mathematically, it is written as;

$\cosh x=\frac{e^x+e^{-x}}{2}$

### Integral of cosh x formula

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

$∫\cosh(x)dx = \sinh x + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. Replace x by 2x to evaluate the integral of cosh(2x).

## How to calculate the integral of cosh(x)?

The integral of cosh 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 cosine by using:

- Derivatives
- Substitution method
- Definite integral

## Integral of coshx by using derivatives

The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. Therefore, we can use the derivative to calculate the integral of a function. Let’s discuss calculating the integral of cosh x by using derivatives.

### Proof of integral of cosh x by using derivatives

Since we know that the integration is the reverse of the derivative. Therefore, we can calculate the integral of cos x by using its derivative. For this, we have to look for some derivatives formulas or a formula that gives cos x as the derivative of any function.

In derivative, we know that,

$\frac{d}{dx}[\sinh(x)] = \cosh(x)$

It means that the derivative of sinh x gives us cosh x. But it has a negative sign. Therefore, to obtain the integral of sine, we have to use it as the integral of cos, that is:

$\frac{d}{dx}[\sinh(x)]=\cosh(x)$

Hence the integral of cos x is equal to the negative of cosh x. It is written as:

$∫\cosh(x)dx = \sinh(x) + c$

## Integral of cosh x by using substitution method

The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of cosh x by using the substitution method.

### Proof of Integral of cosh x by using substitution method

To prove the integral of cosh x by using the substitution formula calculator, suppose that:

$y = \cosh(x)$

Differentiating with respect to x,

$\frac{dy}{dx}= \sinh(x)$

To calculate integral, we can write the above equation as:

$dy =\sinh(x).dx$

By trigonometric identities, we know that sinh x = √cosh²x – 1. Then the above equation becomes,

$dy = \sqrt{\cosh²x – 1}. dx$

Now, substituting the value of sin2 x, such as:

$dy =\sqrt{\cosh²x – 1}.dx$

Multiplying both sides by cos x,

$\frac{\cosh(x)dy}{\sqrt{\cosh²(x) – 1}} = \cosh(x).dx$

Again substitute cosh x = y on the left side.

$\frac{ydy}{\sqrt{\cosh²(x) – 1}} = \cosh(x).dx$

Integrating on both sides by applying integral,

$∫\frac{ydy}{\sqrt{\cosh²(x) – 1}} = ∫\cosh(x)dx$

Let y² - 1 = u.

Then

$2ydy = du\quad \text{or}\quad ydy =\frac{1}{2}du$

Then the above left-hand side integral becomes,

$\frac{1}{2}∫\frac{1}{\sqrt u}du =∫\cosh(x)dx$

$\frac{1}{2}∫u^{-\frac{1}{2}}du =∫\cosh(x)dx$

Since the power rule of integration is

$∫ x^n dx = \frac{x^{n+1}}{n+1} + C$

Therefore, by using this formula we get,

$\frac{1}{2}\left(\frac{u^{\frac{1}{2}}}{(1/2)}\right) + C = ∫\cosh(x)dx$

$u^{\frac{1}{2}} + C = ∫\cosh(x)dx$

Again substituting u = 1 - y², we get

$(1 - y^2)^{\frac{1}{2}} + C = ∫\cosh(x)dx$

And again Substitute y = cosh x here,

$\left(1 - \cosh^2(x)\right)^{\frac{1}{2}} + C = ∫\cosh(x) dx$

$\left(\sinh²(x)\right)^{\frac{1}{2}} + C =∫\cosh(x)dx$

$\sinh(x) + C = ∫\cosh(x) dx$

Hence the integral of cosh x is sinh x. You can also calculate the integral of cosh^2x by using this method.

## Integral of cosh x by using definite integral

The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of cosh x by using the definite integral.

### Proof of integral of cosh x by using definite integral

To compute the integral of cosh x by using a definite integral, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of cosh x from 0 to π. For this, we can write the integral as:

$∫^π_0 \cosh(x)dx = \sinh(x)|^π_0$

Now, substitute the limit in the given function.

$∫^π_0 \cosh(x)dx = \sinh(π) - \sinh(0)$

Since sinh 0 is equal to 0 and sinh π is equal to 0, therefore,

$∫^π_0 \cosh(x)dx = 0$

Which is the calculation of the definite integral of cosh x. Now to calculate the definite integral of cosh x between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$∫^{\frac{π}{2}}_0 \cosh(x)dx = \sinh(x)|^{\frac{π}{2}}_0$

Now,

$∫^{\frac{π}{2}}_0 \cosh(x)dx = sin \frac{π}{2} - sin (0)$

Since sin 0 is equal to 1 and sin π/2 is equal to 1, therefore,

$∫^{\frac{π}{2}}_0 \cosh(x)dx = 1 - 0=1$

Therefore, the definite integral of cosh x is equal to 1.