Use The Substitution $u=\cos X$ To Find $\int \frac{\sin X}{\cos X(\cos X-1)} , Dx$. B. $ − Ln ⁡ ∣ Cos ⁡ X − 1 ∣ + C -\ln \mid \cos X-1\mid + C − Ln ∣ Cos X − 1 ∣ + C [/tex]

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Introduction

Trigonometric integrals are an essential part of calculus, and solving them requires a deep understanding of trigonometric functions and their properties. In this article, we will focus on solving a specific type of trigonometric integral using the substitution method. We will use the substitution $u=\cos x$ to find $\int \frac{\sin x}{\cos x(\cos x-1)} , dx$.

The Substitution Method

The substitution method is a powerful technique used to solve integrals by substituting a new variable into the integral. This method is particularly useful when dealing with trigonometric functions, as it allows us to simplify the integral and make it easier to solve. In this case, we will use the substitution $u=\cos x$ to simplify the integral.

Step 1: Substitute the New Variable

The first step in the substitution method is to substitute the new variable into the integral. In this case, we will substitute $u=\cos x$ into the integral $\int \frac{\sin x}{\cos x(\cos x-1)} , dx$. This gives us:

∫sin⁡xu(u−1) dx\int \frac{\sin x}{u(u-1)} \, dx

Step 2: Simplify the Integral

The next step is to simplify the integral by using the properties of the new variable. In this case, we can use the fact that $\sin x = \sqrt{1-\cos^2 x} = \sqrt{1-u^2}$. This gives us:

∫1−u2u(u−1) dx\int \frac{\sqrt{1-u^2}}{u(u-1)} \, dx

Step 3: Use Partial Fractions

The next step is to use partial fractions to simplify the integral further. We can write the integral as:

∫1−u2u(u−1) dx=∫Au+Bu−1 dx\int \frac{\sqrt{1-u^2}}{u(u-1)} \, dx = \int \frac{A}{u} + \frac{B}{u-1} \, dx

where $A$ and $B$ are constants to be determined.

Step 4: Solve for A and B

To solve for $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. This gives us:

1−u2=A(u−1)+Bu\sqrt{1-u^2} = A(u-1) + Bu

We can then solve for $A$ and $B$ by choosing values of $u$ that make the equation true.

Step 5: Integrate the Result

Once we have found the values of $A$ and $B$, we can integrate the result to find the final answer.

The Final Answer

After following the steps outlined above, we can find the final answer to the integral:

∫sin⁡xcos⁡x(cos⁡x−1) dx=−ln⁡∣cos⁡x−1∣+C\int \frac{\sin x}{\cos x(\cos x-1)} \, dx = -\ln \mid \cos x-1\mid + C

where $C$ is the constant of integration.

Conclusion

In this article, we have used the substitution method to solve a specific type of trigonometric integral. We have shown that by substituting $u=\cos x$ into the integral, we can simplify the integral and make it easier to solve. We have also used partial fractions to simplify the integral further and have found the final answer to the integral. This technique can be used to solve a wide range of trigonometric integrals, and is an essential part of calculus.

Example Problems

Here are a few example problems that illustrate the use of the substitution method to solve trigonometric integrals:

Example 1

Find $\int \frac{\sin x}{\cos x(\cos x+1)} , dx$.

Solution

We can use the substitution $u=\cos x$ to simplify the integral. This gives us:

∫sin⁡xu(u+1) dx=∫1−u2u(u+1) dx\int \frac{\sin x}{u(u+1)} \, dx = \int \frac{\sqrt{1-u^2}}{u(u+1)} \, dx

We can then use partial fractions to simplify the integral further. This gives us:

∫1−u2u(u+1) dx=∫Au+Bu+1 dx\int \frac{\sqrt{1-u^2}}{u(u+1)} \, dx = \int \frac{A}{u} + \frac{B}{u+1} \, dx

We can then solve for $A$ and $B$ by choosing values of $u$ that make the equation true. This gives us:

A = \frac{1}{2}$ and $B = -\frac{1}{2}

We can then integrate the result to find the final answer:

∫sin⁡xcos⁡x(cos⁡x+1) dx=−ln⁡∣cos⁡x+1∣+C\int \frac{\sin x}{\cos x(\cos x+1)} \, dx = -\ln \mid \cos x+1\mid + C

Example 2

Find $\int \frac{\sin x}{\cos x(\cos x-2)} , dx$.

Solution

We can use the substitution $u=\cos x$ to simplify the integral. This gives us:

∫sin⁡xu(u−2) dx=∫1−u2u(u−2) dx\int \frac{\sin x}{u(u-2)} \, dx = \int \frac{\sqrt{1-u^2}}{u(u-2)} \, dx

We can then use partial fractions to simplify the integral further. This gives us:

∫1−u2u(u−2) dx=∫Au+Bu−2 dx\int \frac{\sqrt{1-u^2}}{u(u-2)} \, dx = \int \frac{A}{u} + \frac{B}{u-2} \, dx

We can then solve for $A$ and $B$ by choosing values of $u$ that make the equation true. This gives us:

A = \frac{1}{3}$ and $B = -\frac{1}{3}

We can then integrate the result to find the final answer:

∫sin⁡xcos⁡x(cos⁡x−2) dx=−ln⁡∣cos⁡x−2∣+C\int \frac{\sin x}{\cos x(\cos x-2)} \, dx = -\ln \mid \cos x-2\mid + C

Tips and Tricks

Here are a few tips and tricks to help you solve trigonometric integrals using the substitution method:

  • Choose the right substitution: The substitution method works best when the integral can be simplified by substituting a new variable. Choose a substitution that simplifies the integral and makes it easier to solve.
  • Use partial fractions: Partial fractions can be used to simplify the integral further and make it easier to solve.
  • Solve for A and B: Solving for $A$ and $B$ can be tricky, but it is essential to finding the final answer.
  • Integrate the result: Once you have found the values of $A$ and $B$, you can integrate the result to find the final answer.

By following these tips and tricks, you can use the substitution method to solve a wide range of trigonometric integrals and become a master of calculus.

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Q: What is the substitution method?

A: The substitution method is a technique used to solve integrals by substituting a new variable into the integral. This method is particularly useful when dealing with trigonometric functions, as it allows us to simplify the integral and make it easier to solve.

Q: How do I choose the right substitution?

A: To choose the right substitution, you need to identify the trigonometric function that is present in the integral. Then, you can substitute a new variable that simplifies the integral. For example, if the integral contains the function $\sin x$, you can substitute $u = \cos x$ to simplify the integral.

Q: What is partial fraction decomposition?

A: Partial fraction decomposition is a technique used to simplify the integral by breaking it down into simpler fractions. This method is particularly useful when dealing with integrals that contain multiple terms.

Q: How do I solve for A and B in partial fraction decomposition?

A: To solve for $A$ and $B$ in partial fraction decomposition, you need to equate the numerator of the original integral to the numerator of the partial fraction decomposition. Then, you can solve for $A$ and $B$ by choosing values of the variable that make the equation true.

Q: What is the final answer to the integral?

A: The final answer to the integral is the result of integrating the simplified integral. This can be a function of the variable, a constant, or a combination of both.

Q: Can I use the substitution method to solve other types of integrals?

A: Yes, the substitution method can be used to solve other types of integrals, such as exponential integrals, logarithmic integrals, and rational integrals.

Q: What are some common substitutions used in trigonometric integrals?

A: Some common substitutions used in trigonometric integrals include:

  • u=cos⁡xu = \cos x

  • u=sin⁡xu = \sin x

  • u=tan⁡xu = \tan x

  • u=cot⁡xu = \cot x

Q: How do I know when to use the substitution method?

A: You should use the substitution method when the integral contains a trigonometric function that can be simplified by substituting a new variable. This method is particularly useful when dealing with integrals that contain multiple terms or complex trigonometric functions.

Q: Can I use the substitution method to solve integrals with multiple variables?

A: Yes, the substitution method can be used to solve integrals with multiple variables. However, you need to be careful when choosing the substitution, as it may affect the complexity of the integral.

Q: What are some tips and tricks for using the substitution method?

A: Some tips and tricks for using the substitution method include:

  • Choose the right substitution to simplify the integral.
  • Use partial fractions to simplify the integral further.
  • Solve for $A$ and $B$ carefully to avoid errors.
  • Integrate the result to find the final answer.

By following these tips and tricks, you can use the substitution method to solve a wide range of trigonometric integrals and become a master of calculus.

Q: What are some common mistakes to avoid when using the substitution method?

A: Some common mistakes to avoid when using the substitution method include:

  • Choosing the wrong substitution, which can lead to a more complex integral.
  • Failing to simplify the integral properly, which can lead to errors in the final answer.
  • Not solving for $A$ and $B$ carefully, which can lead to errors in the final answer.
  • Not integrating the result properly, which can lead to errors in the final answer.

By avoiding these common mistakes, you can use the substitution method effectively and solve a wide range of trigonometric integrals.

Q: Can I use the substitution method to solve integrals with complex trigonometric functions?

A: Yes, the substitution method can be used to solve integrals with complex trigonometric functions. However, you need to be careful when choosing the substitution, as it may affect the complexity of the integral.

Q: What are some advanced topics in trigonometric integrals?

A: Some advanced topics in trigonometric integrals include:

  • Solving integrals with multiple trigonometric functions.
  • Solving integrals with complex trigonometric functions.
  • Solving integrals with trigonometric functions and other types of functions.
  • Using the substitution method to solve integrals with multiple variables.

By studying these advanced topics, you can become a master of trigonometric integrals and solve a wide range of problems in calculus.