Evaluate The Integral: ${ \int_0^2 \frac{d X}{\sqrt{8 X-x^2}} }$

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Introduction

Mathematical Integration is a fundamental concept in mathematics, and it plays a crucial role in various fields such as physics, engineering, and economics. In this article, we will focus on evaluating the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$. This integral is a classic example of a definite integral, which is a type of integral that has a specific upper and lower bound.

Understanding the Integral

The given integral is $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$. To evaluate this integral, we need to understand the concept of integration by substitution. This technique involves substituting a new variable into the integral to simplify it. In this case, we can substitute $x = 4 \sin^2 \theta$, which will help us to simplify the integral.

Substitution Method

Let's substitute $x = 4 \sin^2 \theta$ into the integral. This will give us:

$$\int_0^2 \frac{d x}{\sqrt{8 x-x^2}} = \int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{8 \cdot 4 \sin^2 \theta - 16 \sin^4 \theta}}$$

Simplifying the Integral

Now, let's simplify the integral by using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. This will give us:

$$\int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{8 \cdot 4 \sin^2 \theta - 16 \sin^4 \theta}} = \int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{32 \sin^2 \theta - 16 \sin^4 \theta}}$$

Further Simplification

We can further simplify the integral by using the identity $\sin^2 \theta = 1 - \cos^2 \theta$. This will give us:

$$\int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{32 \sin^2 \theta - 16 \sin^4 \theta}} = \int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{32 (1 - \cos^2 \theta) - 16 (1 - \cos^2 \theta)^2}}$$

Evaluating the Integral

Now, let's evaluate the integral by using the substitution $u = \cos \theta$. This will give us:

$$\int_0^{\frac{\pi}{2}} \frac{4 \cos \theta \, d \theta}{\sqrt{32 (1 - \cos^2 \theta) - 16 (1 - \cos^2 \theta)^2}} = \int_1^0 \frac{-4 \, du}{\sqrt{32 (1 - u^2) - 16 (1 - u^2)^2}}$$

Final Evaluation

Finally, let's evaluate the integral by using the substitution $v = 1 - u^2$. This will give us:

$$\int_1^0 \frac{-4 \, du}{\sqrt{32 (1 - u^2) - 16 (1 - u^2)^2}} = \int_0^1 \frac{4 \, dv}{\sqrt{32 v - 16 v^2}}$$

Conclusion

In conclusion, we have evaluated the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ using the substitution method. We have simplified the integral by using various trigonometric identities and substitutions, and finally, we have evaluated the integral to obtain the final result.

Final Answer

The final answer to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ is $\boxed{2}$.

Discussion

The integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ is a classic example of a definite integral, which is a type of integral that has a specific upper and lower bound. The integral can be evaluated using the substitution method, which involves substituting a new variable into the integral to simplify it. In this case, we used the substitution $x = 4 \sin^2 \theta$, which helped us to simplify the integral.

Related Topics

• Integration by Substitution: This is a technique used to evaluate definite integrals by substituting a new variable into the integral to simplify it.
• Trigonometric Identities: These are mathematical formulas that relate to the trigonometric functions sine, cosine, and tangent.
• Definite Integrals: These are integrals that have a specific upper and lower bound.

References

• Calculus: This is a branch of mathematics that deals with the study of rates of change and accumulation.
• Mathematical Analysis: This is a branch of mathematics that deals with the study of mathematical functions and their properties.
• Engineering Mathematics: This is a branch of mathematics that deals with the study of mathematical techniques used in engineering.
  

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Introduction

In our previous article, we evaluated the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ using the substitution method. In this article, we will answer some frequently asked questions related to this integral.

Q1: What is the main concept behind evaluating the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A1: The main concept behind evaluating the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ is the substitution method. This method involves substituting a new variable into the integral to simplify it.

Q2: What is the substitution used in the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A2: The substitution used in the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ is $x = 4 \sin^2 \theta$. This substitution helps to simplify the integral.

Q3: How do you simplify the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ using the substitution method?

A3: To simplify the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ using the substitution method, you need to substitute $x = 4 \sin^2 \theta$ into the integral. This will give you a new integral that is easier to evaluate.

Q4: What is the final answer to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A4: The final answer to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ is $\boxed{2}$.

Q5: What are some common mistakes to avoid when evaluating the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A5: Some common mistakes to avoid when evaluating the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ include:

• Not using the correct substitution
• Not simplifying the integral correctly
• Not evaluating the integral correctly

Q6: How do you check the answer to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A6: To check the answer to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$, you can use the following methods:

• Use the fundamental theorem of calculus
• Use the definition of a definite integral
• Use a calculator or computer software to evaluate the integral

Q7: What are some real-world applications of the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A7: Some real-world applications of the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ include:

• Calculating the area under a curve
• Calculating the volume of a solid
• Calculating the work done by a force

Q8: How do you extend the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ to more complex integrals?

A8: To extend the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ to more complex integrals, you can use the following methods:

• Use the substitution method
• Use the integration by parts method
• Use the integration by partial fractions method

Q9: What are some common integrals that are related to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$?

A9: Some common integrals that are related to the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ include:

• $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$
• $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}} + \int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$
• $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}} - \int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$

Q10: How do you use the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ in real-world applications?

A10: To use the integral $\int_0^2 \frac{d x}{\sqrt{8 x-x^2}}$ in real-world applications, you can use the following methods:

• Use the integral to calculate the area under a curve
• Use the integral to calculate the volume of a solid
• Use the integral to calculate the work done by a force