# What is the Solution for Finding c in the Integral of ArcCos Problem?

• XJellieBX
In summary, the region between y=cos x and the x-axis for x \in [0, \pi/2] is divided into two subregions of equal area by a line y=c. The area of each portion is A_{1}, and the area of the whole region is A_{2}. The region between y=cos x and the x-axis for x \in [0, \pi/2] is divided into two subregions of equal area by a line y=c. The area of each portion is A_{1}, and the area of the whole region is A_{2}. The value of c is found to be .4.
XJellieBX

## Homework Statement

The region between y=cos x and the x-axis for x $$\in$$ [0, $$\pi$$/2] is divided into two subregions of equal area by a line y=c. Find c.

2. The attempt at a solution
First I drew a graph of the region bounded between the function and the x-axis in [0, $$\pi$$/2]. Next I found the total area of the region:
$$A_{1,2}$$ = $$\int^{\pi/2}_{0}$$ cos x dx = sin x |$$^{\pi/2}_{0}$$ = 1

Dividing the total area by 2, gives the area of each portion divided by y=c.

$$A_{1}$$ = 1/2 = $$\int^{c}_{0}$$ $$cos^{-1}$$ (y) dy = y$$cos^{-1}$$ (y) - $$\sqrt{1-y^{2}}$$ $$|^{c}_{0}$$
= c (cos$$^{-1}$$ (c)) - $$\sqrt{1-c^{2}}$$ + 1

--> c ($$cos^{-1}$$ (c)) - $$\sqrt{1-c^{2}}$$ = -1/2

I'm not sure how to solve for c from the equation above. Any input will be appreciated.

Last edited:
XJellieBX said:
--> c ($$cos^{-1}$$ (c)) - $$\sqrt{1-c^{2}}$$ = -1/2

I'm not sure how to solve for c from the equation above. Any input will be appreciated.

I don't think think an exact solution is possible. Try estimating the solution using Newtons Method or any other numerical technique you have learned.

You're not going to be able to solve for an exact value of c using algebraic techniques. The best you can do is to get an approximate value. A simple way to do this is to start with an educated guess, like, say c = 0.4. (The value has to be less than 1/2 because there is more area in the lower part of the region than in the upper part.)

You want to find a value c so that c*cos^(-1)(c) = sqrt(1 - c^2) + 1/2 = 0. c = .4 probably isn't right, but it will give you an idea of what to use for your next try. Keep doing this, with better and better approximations until your successive approximations are in agreement in two, three, or four decimal places, or however precise you want to be.

I had thought it would be just approximating. Thank you =)

## 1. What is the Integral of ArcCos?

The integral of ArcCos is a mathematical operation that calculates the area under the curve of the inverse trigonometric function ArcCos (also known as arccosine) over a given interval. It is denoted by ∫arccos(x)dx and is often used in calculus and other areas of mathematics.

## 2. How is the Integral of ArcCos solved?

The integral of ArcCos can be solved using various methods, such as substitution, integration by parts, or trigonometric identities. The specific method used depends on the complexity of the function and the given interval. It is important to have a good understanding of integration techniques and trigonometric functions to solve this integral.

## 3. What are the common applications of the Integral of ArcCos?

The Integral of ArcCos has various applications in physics, engineering, and other fields of science. It is commonly used to calculate the work done by a force acting along a curved path, as well as in calculating the volume of certain shapes and surfaces. It is also used in solving differential equations and finding the area of irregular regions.

## 4. Are there any special cases when solving the Integral of ArcCos?

Yes, there are special cases when solving the Integral of ArcCos. One such case is when the function inside the integral is a constant, in which case the integral evaluates to a simple formula. Another case is when the function is a trigonometric function with specific values, such as when the function is equal to 1 or 0, making the integral easier to solve.

## 5. What are some tips for solving the Integral of ArcCos?

Some useful tips for solving the Integral of ArcCos include using trigonometric identities and substitution to simplify the integral, paying attention to any special cases, and practicing with different problems to improve problem-solving skills. It is also helpful to have a good understanding of the properties and graphs of inverse trigonometric functions.

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