## Introduction to integral of cos^{-1}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 cubic sine. You will also understand how to compute cos inverse integral by using different integration techniques.

## What is the integral of cos^-1x?

The integral of cos^-1x is an antiderivative of sine function which is equal to x cos^{-1}x – √(1 - x2)+ c. It is also known as the reverse derivative of sine function which is a trigonometric identity. The inverse trigonometric function cos is written as cos^-1x or arccos x. The range of cos inverse lies from 0 to pi and the range lies from -1 to 1.

Our advanced integral calculator allows you to find the integral of cos inverse in just a few steps.

### Integral of cos^{-1}x formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(cos^{-1}x)dx. In mathematical form, the integral of sin^{-1}x is:

$∫\cos^{-1}xdx=x\cos^{-1}x–\sqrt{(1 - x2^)}+ c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. The integral of arccos(x) is used to calculate the unknown angle between two straight line or two curves.

## How to calculate the integral of cos^{-1}x?

The integral of cos^{-1}x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of cosine by using:

- Integration by parts
- Substitution method
- Definite integral

## Integral of cos inverse 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 integral of two functions combined together. Let’s discuss calculating the integral of cos inverse x by using integration by parts.

### Proof of integral of cos^-1x by using integration by parts

Since we know that the function cosine squared x can be written as the product of two functions. Therefore, we can calculate the integral of cos^{-1}x by using integration by parts. For this, suppose that:

$I = \cos^{-1}x = 1.\cos^{-1}x$

Applying the integral we get,

$I = ∫(1.\cos^{-1}x)dx$

Since the method of integration by parts is:

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

Now replacing f(x) and g(x) by cos-1x, we get,

$I = x\cos^{-1}x – ∫\frac{x}{\sqrt{1 – x^2}}dx$

Now multiplying and dividing by 2, we get,

$I = x\cos^{-1}x– \frac{1}{2}∫\frac{-2x}{\sqrt{1 - x^2}}dx$

$I = x\cos^{-1}x –\frac{1}{2}∫(-2x)(1 - x^2)^{-1/2}dx$

$I = x\cos^{-1}x – \frac{1}{2}\left[\frac{(1 - x^2)^{-\frac{1}{2} + 1}}{(\frac{-1}{2} + 1)}\right] + C$

More simplification,

$I = x\cos^{-1}x–\frac{1}{2} \left[\frac{(1 - x^2)^{1/2}}{1/2}\right] + C{2}lt;/p>

$I = x\cos^{-1}x–(1 - x^2)^{1/2} + C$

Hence the derivation of integral of cos inverse x is:

$I=x\cos^{-1}x–\sqrt{1 - x^2} + C$

## Integral of cos^-1x 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 sin inverse by using the substitution method.

### Proof of Integral of cos^-1x by using substitution method

To proof the integral of cos^-1x by using substitution method, the following formulas can be used.

- Assume x = cosθ ⇒ cos-1x = cos-1(cos θ) = θ
- dx = - sinθ dθ
- ∫f(x)g(x)dx = f(x) ∫g(x)dx - ∫[d(f(x))/dx × ∫g(x) dx] dx(Integration by parts rule)
- sin
^{2}θ + cos^{2}θ = 1 ⇒ sinθ = √(1 - cos^{2}θ)

Using the above formulas, we have

$∫\cos^{-1}x dx = -∫\cos^{-1}(\cosθ) \sinθdθ$

$∫\cos^{-1}x dx= - ∫θsinθ dθ$

Now by substituting f(θ) = θ and g(θ) = sinθ in Integration by parts formula

$= - θ∫sinθdθ + ∫[d(θ)/dθ × ∫sinθ dθ]$

$= θ\cosθ – ∫1.\cosθ dθ$

$= θ\cosθ - ∫\cosθ dθ$

$= θ\cos θ – \sinθ + C$

$= θ\cos θ – √(1 - cos2θ) + C$

$= x\cos^{-1}x–\sqrt{1 - x^2} + C$

Hence the integration of cos^-1x is verified by using the substitution method.

## Integral of cos^-1x 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 cos^-1x by using the definite integral finder.

### Proof of integral of cos^-1x by using definite integral

To compute the integral of cos^-1x by using a definite integral, we can use the interval from 0 to 1. Let’s compute the integral of cos^-1x from 0 to 1. Since we know the integral of sine inverse x,

$I = x\cos^{-1}x–\sqrt{1 - x^2} + C$

The indefinite integral of cos^-1x can be written as:

$∫^1_0 \cos^{-1}x dx = [x \cos^{-1}x – \sqrt{1 - x^2}]|^1_0$

Substituting the value of limit we get,

$∫^1_0 \cos^{-1}x dx = [\cos^{-1}1 –\sqrt{1 - 1^2}]–[0\cos^{-1}0–\sqrt{1 - 0}]$

Since cos inverse 1 is equal to 0, then,

$∫^1_0 \cos^{-1}x dx = 0 – (– 1)$

Therefore, the integral of cos-1x from 0 to 1 is

$∫^1_0 \cos^{-1}x dx = 1$

Which is the calculation of the definite integral of cos^-1x. The definite integral is a way to find the area bounded by two points. You can also use our area under the curve calculator to find area under a curve.