## Introduction to Integral of cos sqrt x

In calculus, the integral calculator follows 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 cos(√x) integration by using different integration techniques.

## What is the integration of cos√x?

The integral of cos(√x) is an antiderivative of the cos sqrt function which is equal to 2{√x.sin(√x) – cos(√x)}. It is also known as the reverse derivative of the cos(sqrt x) function which is a trigonometric identity. The cosine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:

cos = adjacent side/hypotenuse

The integral of cos sqrt x is a complex integral in calculus because the integrand contains a trigonometric function cos with a square root of angle x. It helps to evaluate many different integral problems in calculus.

### Integral of cos(√x) formula

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

$\int \cos(\sqrt x)dx = 2[\sqrt x.\sin(\sqrt x) + cos(\sqrt x)]$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. Replacing cos(√x) with cos(e^x) in the above formula will give the integral of cos(ex).

## How to calculate the integral of cos(√x)?

The integral of cos √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:

- Integration by Parts
- Substitution method
- Definite integral

## Integral of cos √x by integration by parts

The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving the integral of two functions combined together. Let’s discuss calculating the integral of cos √x by using integration by parts.

### Proof of integral of cos x by using integration by parts

Since we know that the function cosine √x can be written as the product of two functions. Therefore, we can calculate the integral of cos √x by using integration by parts. For this, suppose that:

$I = \int \cos(\sqrt x)dx$

Suppose that,

$u = \sqrt x \quad\text{and}\quad du = \frac{1}{2\sqrt x}.dx$

Or,

$du = \frac{1}{2u}dx\quad\text{or}\quad2u.du = dx$

Then,

$I = \int 2u.\cos udx$

Now, using the integration by parts rule,

$\int [f(x).g(x)]dx = f(x)\int g(x)dx – \int [f’(x).\int g(x)dx]dx$

Then, apply this to the above integral and consider f(x) =u and g(x) = cos u.

Therefore,

$I = 2u.\sin u – 2\int \sin u.du$

Moreover,

$I = 2u.\sin u + 2\cos u$

Now substituting the value of u, we get,

$I = 2[\sqrt x.\sin(\sqrt x) + \cos(\sqrt x)]$

Hence the integration of cos√x by using integration by parts calculator.

## Integral of cos root 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 cos(sqrt(x)) by using the substitution method.

### Proof of cos(sqrt(x)) integral by using substitution method

To prove the cos√x integral by using the substitution method, suppose that:

$y = \cos(\sqrt x)$

Differentiating with respect to x,

$\frac{dy}{dx}=-\sin(\sqrt x).\left(\frac{1}{2\sqrt x}\right)$

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

$dy = -\sin(\sqrt x).\left(\frac{1}{2\sqrt x}\right)dx$

By trigonometric identities, we know that sin √x = √1 - cos²√x. Then the above equation becomes,

$dy = -\frac{1}{2\sqrt x}\sqrt{1 - \cos^2\sqrt x}dx$

Now, substituting the value of cos2 √x, such as:

$dy = -\frac{1}{2\sqrt x}\sqrt{1 – y^2}.dx$

Multiplying both sides by cos x,

$-\frac{\cos(\sqrt x)dy}{2\sqrt x\sqrt{1 - y²}} = \cos(\sqrt x).dx$

Again substitute cos x = y on the left side.

$-\frac{ydy}{2\sqrt x\sqrt{1 - y^2}}=\cos(\sqrt x).dx$

Integrating on both sides by applying integral,

$-\int \frac{ydy}{2\sqrt x\sqrt{1 - y^2}} =\int \cos(\sqrt x)dx$

Now by using the u-substitution formula,

$\text{Let}\quad 1 - y^2 = u$

Then

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

Then the above left-hand side integral becomes,

$\int \frac{\sqrt x}{\sqrt u}du =\int \cos(\sqrt x)dx$

$\int \sqrt x u^-{1/2}du =\int \cos(\sqrt x)dx{2}lt;/p>

Since the power rule of integration is,

$\int x^n dx = \frac{x^{n+1}}{(n+1)} + C$

Therefore, by using this formula we get

$\frac{\sqrt x.u^{1/2}}{(1/2)} + C =\int \cos(\sqrt x) dx$

$2\sqrt x\times\sqrt u + C =\int \cos(\sqrt x) dx$

Again substituting u = 1 - y², we get

$2\sqrt x (1 - y^2)^{1/2} + C =\int \cos(\sqrt x)dx$

And again Substitute y = cos √x here,

$2\sqrt x (1 - \cos^2(\sqrt x))^{1/2} + C =\int \cos(\sqrt x)dx$

$2\sqrt x (\sin^2(\sqrt x))^{1/2} + C =\int \cos(\sqrt x)dx$

$2\sqrt x.\sin(\sqrt x)+C =\int \cos(\sqrt x)dx$

Hence the integral of cos √x is 2√x.sin √x. This method is also applicable to calculate the integral of cos(x^2).

## Integration of cos√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:

$\int^b_a f(x) dx = F(b) – F(a)$

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

### Proof of cos√x integral by using definite integral

To compute the integral of cos(sqrt(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 cos x from 0 to π. For this, we can write the integral as:

$\int^\pi_0 \cos(\sqrt x)dx = 2[\sqrt x.\sin(\sqrt x) + \cos(\sqrt x)]|^\pi_0$

Now, substitute the limit in the given function.

$\int^\pi_0 \cos(\sqrt x)dx = 2[\sqrt π.\sin(\sqrt π) + \cos(\sqrt π)] – 2[0.\sin 0 + \cos 0]$

Since sin 0 is equal to 0 and sin π is equal to 0, therefore, after solving the above integral we get,

$\int^\pi_0 \cos(\sqrt x)dx = 1.0725$

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

$\int^{\frac{\pi}{2}}_0 \cos(\sqrt x)dx = 2[\sqrt x.\sin(\sqrt x) + \cos(\sqrt x)]|^{\frac{\pi}{2}}_0$

Now,

$\int^{\frac{\pi}{2}}_0 \cos(\sqrt x)dx = 2\left[\sqrt{\frac{π}{2}}.\sin\sqrt{\frac{π}{2}} + \cos\sqrt{\frac{π}{2}}\right] – 2[0.\sin 0 + \cos 0]$

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

$\int^{\frac{\pi}{2}}_0 \cos(\sqrt x)dx = 1 - 0=1$

Therefore, the definite integral of cos root x is equal to 1. The definite integral also helps to calculate the area under a curve.