## Introduction to integral of cos x/sin^{2}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/sin2x integral by using different integration techniques.

## What is the integral of cosx/sin^2x?

The integral of cos x/sin^{2}x is an antiderivative of sine function which is equal to – csc x +c. It is also known as the reverse derivative of cos x/sin^2x 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 cosine function is the ratio of the adjacent side to the hypotenuse of a triangle which is written as:

Cos = adjacent side/hypotenuse

The integral of cos x by sin^2x is a fractional integral in calculus. It allows us to understand how to solve integrals with fractions. Such as the integral of cos x/1+sin x.

### Integral of cos x/sin^{2}x formula

The formula of the integral of cos x by sin square x contains the integral sign, coefficient of integration, and the function as sine. It is denoted by ∫(cos x/sin^{2}x)dx. In mathematical form, the integral of cos x/sin^{2}x is:

$\int \frac{\cos x}{\sin^2x}dx = - \csc x + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. Using the above formula, we can also calculate the integral of sin x/cos^2x using different integration approaches.

## How to calculate the integral of cos (x)/sin^{2}(x)?

The integral of cos x/sin^{2}x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using:

- Integration by parts
- Substitution method
- Definite integral

## Integral of cos x/sin^{2} x by using 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 integration of cos x/sin^{2} x by using integration by parts.

### Proof of integral of cosx/sin^2x by using integration by parts

Since we know that the function sine and cosine squared can be written as the product of two functions. Therefore, we can calculate the integral of sin^2x by using integration by parts. For this, suppose that:

$I = \frac{\cos x}{\sin^2x}$

Applying the integral we get,

$I = \int\frac{\cos x}{\sin^2x}dx$

Since sec x = 1/cos x, then the above integral can also be written as:

$I = \int (\cos x.\csc^2x)dx$

Since the method of integration by parts is:

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

Now replacing f(x) and g(x) by sin x and cos^{2}x, we get,

$I = \sin x.\csc^2x + 2\int [\csc^2x.\cot x.\sin x]dx$

It can be written as:

$I = \sin x.\csc^2x + 2\int [\csc^2x.\cos x]dx$

Since we know that;

$I = \int (\cos x.\csc^2 x)dx=\int \frac{\cos x}{\sin^2 x}dx$

Therefore,

$I = \sin x.\csc^2 x + 2I$

Or,

$I = -\sin x.\csc^2x + c$

Moreover,

$I = - \csc x + c$

Hence the derivation of integral of sin x/cos^{2} x is,

$I = -\csc x + c$

Our by-parts integration tool can also assist you in calculating different integrals by treating them as a product of two functions.

## Integral of cos x/sin2 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/sin^{2}x by using the substitution method.

### Proof of Integral of cos x/sin^{2}x by using substitution method

The method of substitution calculator involves two types, trigonometric and u-substitution. To prove the integral of cosx/sin^2x by using the substitution method, suppose that:

$I = \int \frac{\cos x}{\sin^2 x}dx$

Suppose that, u = sin x and du = cos xdx, then the above integral will become,

$I = \int \frac{1}{u^2}du$

By integrating with respect to u we get;

$I = \frac{u^{-2+1}}{-2+1} + c$

Now substituting the value of u back here,

$I = - u^{-1} + c$

Then,

$I = -\frac {1}{u} + c$

Now substituting the value of u here.

$I = -\frac {1}{\sin x} + c$

Or,

$I = - \csc x +c$

Hence the integral of cos x/sin2 x is equal to sec x. Also, use our trigonometric substitution calculator to solve nonlinear integrals.

## Integral of cos x/sin2 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 sin x by using the definite integral.

### Proof of integral of cos x/sin2 x by using definite integral

To compute the integral of cos x/sin2x by using a definite integral tool, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of cos x/sin2 x from 0 to π. For this we can write the integral as:

$\int^\pi_0 \frac {\cos x}{\sin^2x} dx = - \csc x|^\pi_0$

Now, substituting the limit in the given function.

$\int^\pi_0 \frac {\cos x}{\sin^2x} dx = - \csc (\pi) + \csc (0)$

Since csc 0 is equal to 0 and sec π is equal to 0, therefore,

$\int^\pi_0 \frac {\cos x}{\sin^2x} dx = 0$

Which is the calculation of the definite integral of cos x/sin2 x. Now to calculate the integral of sin x/cos2 x between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$\int^{\frac{\pi}{2}}_0 \frac {\cos x}{\sin^2x} dx = - \csc x|^{\frac{\pi}{2}}_0$

Now,

$\int^{\frac{\pi}{2}}_0 \frac {\cos x}{\sin^2x} dx = - \csc \frac{\pi}{2} + csc (0)$

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

$\int^{\frac{\pi}{2}}_0 \frac {\cos x}{\sin^2x} dx = 0 + 1=1$

Therefore, the definite integral of cosx/sin^2x is equal to 1.