How To Calculate ∫ 0 Π / 2 Arccos ⁡ ( Cos ⁡ X 1 + Cos ⁡ X ) D X \int_{0}^{\pi/2} \arccos\left(\frac{\cos X}{1 + \cos X}\right)\,\mathrm Dx ∫ 0 Π /2 ​ Arccos ( 1 + C O S X C O S X ​ ) D X

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Introduction

Calculating definite integrals can be a challenging task, especially when they involve trigonometric functions. In this article, we will explore the process of calculating the integral $\int_{0}^{\pi/2} \arccos\left(\frac{\cos x}{1 + \cos x}\right)\,\mathrm dx$. This integral is a variation of Coxeter's integral, and we will use integration by parts to evaluate it.

Background

The integral in question is a definite integral, meaning it has a specific upper and lower bound. In this case, the lower bound is 0 and the upper bound is $\frac{\pi}{2}$. The integrand is the arccosine of the expression $\frac{\cos x}{1 + \cos x}$.

Integration by Parts

To evaluate this integral, we will use integration by parts. This technique involves differentiating one function and integrating the other. In this case, we will let $u = \arccos\left(\frac{\cos x}{1 + \cos x}\right)$ and $dv = \mathrm dx$.

Step 1: Differentiate u

To differentiate $u$, we will use the chain rule. Let $v = \frac{\cos x}{1 + \cos x}$. Then, $\frac{\mathrm du}{\mathrm dx} = -\frac{1}{\sqrt{1-v^2}} \frac{\mathrm dv}{\mathrm dx}$.

Step 3: Integrate dv

To integrate $dv$, we simply have $\int \mathrm dx = x + C$.

Step 4: Apply Integration by Parts

Now, we can apply integration by parts. We have:

$$\begin{aligned} I &= \int_{0}^{\pi/2} \arccos\left(\frac{\cos x}{1 + \cos x}\right)\,\mathrm dx \\ &= \left[ x \arccos\left(\frac{\cos x}{1 + \cos x}\right) \right]_{0}^{\pi/2} - \int_{0}^{\pi/2} x \frac{\mathrm du}{\mathrm dx} \,\mathrm dx \end{aligned}$$

Step 5: Evaluate the First Term

To evaluate the first term, we simply substitute the upper and lower bounds:

$$\begin{aligned} \left[ x \arccos\left(\frac{\cos x}{1 + \cos x}\right) \right]_{0}^{\pi/2} &= \frac{\pi}{2} \arccos(0) - 0 \arccos(0) \\ &= \frac{\pi}{2} \cdot \frac{\pi}{2} \\ &= \frac{\pi^2}{4} \end{aligned}$$

Step 6: Evaluate the Second Term

To evaluate the second term, we need to find $\int_{0}^{\pi/2} x \frac{\mathrm du}{\mathrm dx} \,\mathrm dx$. We can do this by substituting the expression for $\frac{\mathrm du}{\mathrm dx}$:

\begin{aligned} \int_{0}^{\pi/2} x \fracmathrm du}{\mathrm dx} \,\mathrm dx &= \int_{0}^{\pi/2} x \left( -\frac{1}{\sqrt{1-v^2}} \frac{\mathrm dv}{\mathrm dx} \right) \,\mathrm dx \\ &= -\int_{0}^{\pi/2} x \left( -\frac{1}{\sqrt{1-v^2}} \frac{\mathrm dv}{\mathrm dx} \right) \,\mathrm dx \end{aligned}

Step 7: Simplify the Integral

To simplify the integral, we can substitute $v = \frac{\cos x}{1 + \cos x}$. Then, $\frac{\mathrm dv}{\mathrm dx} = -\frac{\sin x}{(1 + \cos x)^2}$.

Step 8: Evaluate the Integral

Now, we can evaluate the integral:

$$\begin{aligned} -\int_{0}^{\pi/2} x \left( -\frac{1}{\sqrt{1-v^2}} \frac{\mathrm dv}{\mathrm dx} \right) \,\mathrm dx &= \int_{0}^{\pi/2} x \left( \frac{1}{\sqrt{1-v^2}} \frac{\mathrm dv}{\mathrm dx} \right) \,\mathrm dx \\ &= \int_{0}^{\pi/2} x \left( \frac{1}{\sqrt{1-v^2}} \left( -\frac{\sin x}{(1 + \cos x)^2} \right) \right) \,\mathrm dx \\ &= -\int_{0}^{\pi/2} \frac{x \sin x}{(1 + \cos x)^2 \sqrt{1-v^2}} \,\mathrm dx \end{aligned}$$

Step 9: Simplify the Integral

To simplify the integral, we can substitute $v = \frac{\cos x}{1 + \cos x}$. Then, $\sqrt{1-v^2} = \sqrt{1-\frac{\cos^2 x}{(1 + \cos x)^2}} = \sqrt{\frac{1 + \cos x}{(1 + \cos x)^2}} = \frac{1}{1 + \cos x}$.

Step 10: Evaluate the Integral

Now, we can evaluate the integral:

$$\begin{aligned} -\int_{0}^{\pi/2} \frac{x \sin x}{(1 + \cos x)^2 \sqrt{1-v^2}} \,\mathrm dx &= -\int_{0}^{\pi/2} \frac{x \sin x}{(1 + \cos x)^2} \frac{1}{\frac{1}{1 + \cos x}} \,\mathrm dx \\ &= -\int_{0}^{\pi/2} \frac{x \sin x}{1 + \cos x} \,\mathrm dx \end{aligned}$$

Step 11: Simplify the Integral

To simplify the integral, we can substitute $u = \cos x$. Then, $\mathrm du = -\sin x \,\mathrm dx$.

Step 12: Evaluate the Integral

Now, we can evaluate the integral:

$$\begin{aligned} -\int_{0}^{\pi/2} \frac{x \sin x}{1 + \cos x} \,\mathrm dx &= \int_{}^{0} \frac{-u}{1 + u} \,\mathrm du \\ &= \int_{0}^{1} \frac{u}{1 + u} \,\mathrm du \end{aligned}$$

Step 13: Evaluate the Integral

To evaluate the integral, we can use partial fractions:

$$\begin{aligned} \int_{0}^{1} \frac{u}{1 + u} \,\mathrm du &= \int_{0}^{1} \left( 1 - \frac{1}{1 + u} \right) \,\mathrm du \\ &= \left[ u - \ln(1 + u) \right]_{0}^{1} \\ &= 1 - \ln 2 \end{aligned}$$

Step 14: Combine the Results

Now, we can combine the results:

$$\begin{aligned} I &= \frac{\pi^2}{4} - \left( 1 - \ln 2 \right) \\ &= \frac{\pi^2}{4} - 1 + \ln 2 \end{aligned}$$

The final answer is: $\boxed{\frac{\pi^2}{4} - 1 + \ln 2}$

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Introduction

In our previous article, we explored the process of calculating the integral $\int_{0}^{\pi/2} \arccos\left(\frac{\cos x}{1 + \cos x}\right)\,\mathrm dx$. This integral is a variation of Coxeter's integral, and we used integration by parts to evaluate it. In this article, we will answer some common questions related to this integral.

Q: What is the significance of this integral?

A: This integral is significant because it is a variation of Coxeter's integral, which is a well-known integral in mathematics. The evaluation of this integral provides insight into the properties of trigonometric functions and their behavior in definite integrals.

Q: What is the main challenge in evaluating this integral?

A: The main challenge in evaluating this integral is the complexity of the integrand, which involves the arccosine function and trigonometric functions. The use of integration by parts and substitution techniques is necessary to simplify the integral and evaluate it.

Q: Can this integral be evaluated using other methods?

A: Yes, this integral can be evaluated using other methods, such as the use of trigonometric identities and the properties of definite integrals. However, the method of integration by parts is a straightforward and efficient way to evaluate this integral.

Q: What is the final answer to this integral?

A: The final answer to this integral is $\frac{\pi^2}{4} - 1 + \ln 2$.

Q: How does this integral relate to other mathematical concepts?

A: This integral is related to other mathematical concepts, such as trigonometry, calculus, and number theory. The evaluation of this integral provides insight into the properties of trigonometric functions and their behavior in definite integrals, which is essential in many mathematical and scientific applications.

Q: Can this integral be used in real-world applications?

A: Yes, this integral can be used in real-world applications, such as in the fields of physics, engineering, and computer science. The evaluation of this integral provides insight into the properties of trigonometric functions and their behavior in definite integrals, which is essential in many mathematical and scientific applications.

Q: What are some common mistakes to avoid when evaluating this integral?

A: Some common mistakes to avoid when evaluating this integral include:

• Not using the correct method of integration, such as integration by parts or substitution.
• Not simplifying the integral correctly, which can lead to incorrect results.
• Not checking the final answer for accuracy, which can lead to errors.

Q: How can I apply this integral in my own work?

A: You can apply this integral in your own work by using it as a tool to evaluate other definite integrals that involve trigonometric functions. The evaluation of this integral provides insight into the properties of trigonometric functions and their behavior in definite integrals, which is essential in many mathematical and scientific applications.

Q: What are some resources for further learning on this topic?

A: Some resources for further learning on this topic include:

• Textbooks on calculus and trigonometry.
• resources, such as Khan Academy and Wolfram Alpha.
• Research papers and articles on the topic of definite integrals and trigonometric functions.

The final answer is: $\boxed{\frac{\pi^2}{4} - 1 + \ln 2}$