## Introduction to the integral of cos x/1+sin 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 cos x by 1+sin x integral by using different integration techniques.

## What is the integral of cos x/1+sin x?

The integral of cos x/1+sin x is an antiderivative of the function cos x/1+sin x which is equal to ln|1+sin x|+c. It is also known as the reverse derivative of the cos x/1+sin x function which is 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

The integral of cos x by 1+sin x is a fractional integral in calculus. It involves a fraction of two functions cos x and 1+sin x which allows us to solve different integration problems containing fractions.

### Integral of cos x/1+sin 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 ∫(cos x/1+sin x)dx. In mathematical form, the integral of sin2x/sin x is:

$\int \left(\frac{\cos x}{1+sin x}\right)dx = \ln|1+ \sin x| + c{2}lt;/p>

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. The above formula can be used to solve the integral of cos x/sin^2x by using different integration techniques such as integration by parts rule, substitution method etc.

## How to calculate the integral of cos x/1+sin x?

The integral of cos x/ 1+sin 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 cos x/1+sin x by using:

- Substitution method
- Definite integral

## Integral of cos x/1+sin 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 x/1+sin x by using the substitution method.

### Proof of Integral of cos x/1+sin x by using substitution method

To prove the integral of sin2x/1+cos2x by using the substitution method, suppose that:

$y=\frac{\cos x}{1+\sin x}{2}lt;/p>

Applying integral,

$I =\int \frac{\cos x}{1+sin x}dx

Therefore assume that

u = 1+sin x and du = cos x dx

Using this substitution in the integral,

$I = \int \frac{du}{u}{2}lt;/p>

Since,

$\frac{d}{dx}(u) = 1{2}lt;/p>

Now integrating with respect to u we get,

$I = \ln|u| + c{2}lt;/p>

Now substituting the value of u here,

$I = \ln|1+\sin x| + c{2}lt;/p>

Hence we have verified the integral of cos x/1+sin x by using the substitution method. We can also find it by using the following alternative method.

$I =\int \left(\frac{\cos x}{1+\sin x}\right)dx{2}lt;/p>

Multiplying and dividing by 1 – sin x, we get,

$I =\int \left(\frac{\cos x(1 – \sin x)}{(1 – \sin^2x)}\right).dx{2}lt;/p>

Since 1 – sin^{2}x = cos^{2}x,

$I =\int \left(\frac{\cos x(1 – \sin x)}{cos^2x}\right).dx{2}lt;/p>

Or,

$I = \int \left(\frac{1 – \sin x}{\cos x}\right)dx{2}lt;/p>

After more simplification,

$I = \int (\sec x – \tan x).dx{2}lt;/p>

Integrating with respect to x,

$I = \ln |\sec x + \tan x| - \ln |\sec x|{2}lt;/p>

Now, we can write it as,

$I = \ln \left|\frac{\sec x + \tan x}{\sec x}\left|{2}lt;/p>

Or,

$I = \ln|1+ \tan x| + c{2}lt;/p>

Using the substitution method, the integral of sin x/cos^2x can also be calculated easily by using different substitutions.

## Integral of cos x/1+sin 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){2}lt;/p>

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

### Proof of integral of cos x/1 + sin x by using definite integral

To compute the integral of cos x/1+sin 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/1+sin x from 0 to π. For this, we can write the integral as:

$\int^\pi_0 \left(\frac{\cos x}{1 + sin x}\right)dx = \ln|1+\sin x||^\pi_0{2}lt;/p>

Now, substitute the limit in the given function.

$\int^\pi_0 \left(\frac{\cos x}{1 + sin x}\right)dx = \ln|1+\sin (\pi)| - \ln|1+\sin (0)|{2}lt;/p>

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

$\int^\pi_0 \frac{\cos x}{1 + \sin x}dx = ln|1|{2}lt;/p>

Which is the calculation of the definite integral of cos x/1+sin x. Now to calculate the integral of cos x/1+sin x between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$\int^\frac{\pi}{2}_0 \frac{\cos x}{1 + \sin x}dx = \ln|1 + \sin x|^\frac{\pi}{2}_0{2}lt;/p>

Now, substitute the limit in the given function.

$\int^\frac{\pi}{2}_0 \frac{\cos x}{1 + \sin x}dx = \ln|1+\sin \frac{\pi}{2}| +\ln|1+\sin 0|{2}lt;/p>

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

$\int^\frac{\pi}{2}_0 \frac{\cos x}{1 + \sin x}dx = \ln(2) – \ln(1) = \ln2

Therefore, the definite integral of cos x/1+sin x is equal to ln2.