Evaluate The Integral: ∫ Cot ⁡ 2 X − 1 Cot ⁡ X D X \int \frac{\cot^2 X - 1}{\cot X} \, Dx ∫ C O T X C O T 2 X − 1 ​ D X

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Introduction

In this article, we will delve into the world of calculus and explore the process of evaluating a specific integral. The given integral is ∫cot⁡2x−1cot⁡x dx\int \frac{\cot^2 x - 1}{\cot x} \, dx. This type of problem is commonly encountered in mathematics, particularly in the study of trigonometric functions and their applications in calculus. Our goal is to simplify the given expression and arrive at a final solution.

Step 1: Simplify the Integral Expression

To begin, let's simplify the given integral expression by using the properties of trigonometric functions. We can rewrite the expression as follows:

∫cot⁡2x−1cot⁡x dx=∫cot⁡2xcot⁡x dx−∫1cot⁡x dx\int \frac{\cot^2 x - 1}{\cot x} \, dx = \int \frac{\cot^2 x}{\cot x} \, dx - \int \frac{1}{\cot x} \, dx

Step 2: Apply Trigonometric Identities

Now, let's apply some trigonometric identities to simplify the expression further. We know that cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}, so we can rewrite the expression as follows:

∫cot⁡2xcot⁡x dx−∫1cot⁡x dx=∫cos⁡xsin⁡x dx−∫sin⁡xcos⁡x dx\int \frac{\cot^2 x}{\cot x} \, dx - \int \frac{1}{\cot x} \, dx = \int \frac{\cos x}{\sin x} \, dx - \int \frac{\sin x}{\cos x} \, dx

Step 3: Use the Quotient Rule

Next, let's use the quotient rule to simplify the expression. The quotient rule states that if we have an expression of the form f(x)g(x)\frac{f(x)}{g(x)}, then the derivative is given by:

ddx(f(x)g(x))=g(x)f′(x)−f(x)g′(x)[g(x)]2\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}

In our case, we have:

ddx(cos⁡xsin⁡x)=sin⁡x(−sin⁡x)−cos⁡x(cos⁡x)sin⁡2x\frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) = \frac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2 x}

Step 4: Simplify the Expression

Now, let's simplify the expression by canceling out the common terms:

sin⁡x(−sin⁡x)−cos⁡x(cos⁡x)sin⁡2x=−sin⁡2x+cos⁡2xsin⁡2x\frac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2 x} = -\frac{\sin^2 x + \cos^2 x}{\sin^2 x}

Step 5: Use the Pythagorean Identity

Next, let's use the Pythagorean identity to simplify the expression further. The Pythagorean identity states that:

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

So, we can rewrite the expression as follows:

−sin⁡2x+cos⁡2xsin⁡2x=−1sin⁡2x-\frac{\sin^2 x + \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}

Step 6: Integrate the Expression

Now, let's integrate the expression:

∫−1sin⁡2x dx=−∫csc⁡2x dx\int -\frac{1}{\sin^2 x} \, dx = -\int \csc^2 x \, dx

Step 7: Evaluate the Integral

Finally, let's evaluate the integral:

−∫csc⁡2x dx=−(−cot⁡x)+C-\int \csc^2 x \, dx = -(-\cot x) + C

Conclusion

In conclusion, we have successfully evaluated the given integral:

∫cot⁡2x−1cot⁡x dx=−(−cot⁡x)+C\int \frac{\cot^2 x - 1}{\cot x} \, dx = -(-\cot x) + C

This result can be verified by differentiating the final expression and checking that it matches the original integral.

Final Answer

The final answer is: −cot⁡x+C\boxed{-\cot x + C}

Introduction

In our previous article, we evaluated the integral ∫cot⁡2x−1cot⁡x dx\int \frac{\cot^2 x - 1}{\cot x} \, dx. In this article, we will provide a Q&A section to help clarify any doubts and provide additional information on the topic.

Q: What is the cotangent function?

A: The cotangent function, denoted by cot⁡x\cot x, is a trigonometric function that is defined as the ratio of the adjacent side to the opposite side in a right triangle. It is the reciprocal of the tangent function.

Q: How do I simplify the integral expression?

A: To simplify the integral expression, we can use the properties of trigonometric functions, such as the Pythagorean identity, to rewrite the expression in a more manageable form. We can also use the quotient rule to simplify the expression.

Q: What is the quotient rule?

A: The quotient rule is a rule in calculus that is used to find the derivative of a quotient of two functions. It states that if we have an expression of the form f(x)g(x)\frac{f(x)}{g(x)}, then the derivative is given by:

ddx(f(x)g(x))=g(x)f′(x)−f(x)g′(x)[g(x)]2\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}

Q: How do I use the Pythagorean identity?

A: The Pythagorean identity states that:

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

We can use this identity to simplify the expression by canceling out the common terms.

Q: What is the final answer?

A: The final answer is:

∫cot⁡2x−1cot⁡x dx=−(−cot⁡x)+C\int \frac{\cot^2 x - 1}{\cot x} \, dx = -(-\cot x) + C

This result can be verified by differentiating the final expression and checking that it matches the original integral.

Q: What is the significance of the constant C?

A: The constant C is an arbitrary constant that is added to the final answer. It represents the fact that the integral is not uniquely determined, and that there are many possible solutions.

Q: How do I apply this result to real-world problems?

A: This result can be applied to a variety of real-world problems, such as:

  • Calculating the area under a curve
  • Finding the volume of a solid
  • Solving optimization problems

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

A: Some common mistakes to avoid when evaluating integrals include:

  • Failing to simplify the expression
  • Using the wrong trigonometric identity
  • Forgetting to add the constant C

Conclusion

In conclusion, we have provided a Q&A section to help clarify any doubts and provide additional information on the topic of evaluating the integral ∫cot⁡2x−1cot⁡x dx\int \frac{\cot^2 x - 1}{\cot x} \, dx. We hope that this article has been helpful in providing a better understanding of the subject.

Final Answer

The final answer is: −cot⁡x+C\boxed{-\cot x + C}