Problem 47
Question
Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(-3 \csc ^{2} x\right) d x$$
Step-by-Step Solution
Verified Answer
The antiderivative is \\(-3 \cot x + C\\).
1Step 1: Identify the Integral
We need to find the antiderivative of \(\int -3 \csc^2 x \, dx\). The integrand is a trigonometric function multiplied by a constant.
2Step 2: Use the Known Antiderivative Formula
Recall that the integral of \(-\csc^2 x\) is \(\cot x\). Therefore, \( \int -\csc^2 x \, dx = \cot x + C\), where \(C\) is the constant of integration.
3Step 3: Apply the Constant Factor
Since there is a factor of \(-3\), multiply the antiderivative \(\cot x\) by \(-3\). So \(\int -3 \csc^2 x \, dx = -3 \cot x + C\).
4Step 4: Differentiate to Check the Solution
Differentiate \(-3 \cot x + C\). The derivative of \(-3 \cot x\) is \(3 \csc^2 x\) by using the derivative of \(\cot x\), which is \(-\csc^2 x\). The derivative of \(C\) is zero. Thus, the derivative is indeed \(-3 \csc^2 x\), confirming our solution.
Key Concepts
Understanding the Cosecant (csc) FunctionTrigonometric Integration TechniquesRole of the Constant of Integration
Understanding the Cosecant (csc) Function
The cosecant function, denoted as \( \csc x \), is a trigonometric function that is the reciprocal of the sine function. In other words, \( \csc x = \frac{1}{\sin x} \). This function is essential in trigonometry because it provides another perspective on relationships within right triangles and in the unit circle.
Knowing these properties becomes handy, especially in calculus when you integrate or differentiate trigonometric functions involving \( \csc x \). For example, the antiderivative of \( -\csc^2 x \) is \( \cot x \). This knowledge directly applies to solving the given integral!
- Reciprocal Identity: \( \csc x = 1/\sin x \)
- Range: \( y \} \) where \( y \) is greater than 1 or less than -1.
- Period: Like sine and cosine, \( \csc x \) repeats every \( 2\pi \).
- Domain: \( x eq k\pi \), where \( k \) is an integer because \( \sin x = 0 \) at these values.
Knowing these properties becomes handy, especially in calculus when you integrate or differentiate trigonometric functions involving \( \csc x \). For example, the antiderivative of \( -\csc^2 x \) is \( \cot x \). This knowledge directly applies to solving the given integral!
Trigonometric Integration Techniques
In calculus, trigonometric integration involves finding the antiderivatives of trigonometric functions. This is a vital technique, often encountered in problems requiring the integration of more complex expressions.
One fundamental approach is to use known antiderivatives of basic trigonometric functions. These known forms simplify the integration process. For example:
This last one is particularly useful for the problem at hand. By recognizing that \( -3 \csc^2 x \) leverages the basic antiderivative \( -\csc^2 x \), we can easily solve for \( \int -3 \csc^2 x \, dx \) by multiplying the antiderivative of \( -\csc^2 x \) by the constant factor \(-3\). This results directly in \(-3 \cot x + C\).Not forgetting, when differentiating back to check, the derivative of \( \cot x \) helps confirm the correctness of the integration solution.
One fundamental approach is to use known antiderivatives of basic trigonometric functions. These known forms simplify the integration process. For example:
- \( \int \sin x \, dx = -\cos x + C \)
- \( \int \cos x \, dx = \sin x + C \)
- \( \int -\csc^2 x \, dx = \cot x + C \)
This last one is particularly useful for the problem at hand. By recognizing that \( -3 \csc^2 x \) leverages the basic antiderivative \( -\csc^2 x \), we can easily solve for \( \int -3 \csc^2 x \, dx \) by multiplying the antiderivative of \( -\csc^2 x \) by the constant factor \(-3\). This results directly in \(-3 \cot x + C\).Not forgetting, when differentiating back to check, the derivative of \( \cot x \) helps confirm the correctness of the integration solution.
Role of the Constant of Integration
When calculating an antiderivative, you encounter the mysterious letter \( C \), known as the constant of integration. This constant is crucial because indefinite integrals do not have a unique solution. Instead, they represent a family of functions that differ by a constant.
This arises because the differentiation of a constant is zero. So, when checking your antiderivative by differentiation, the constant \( C \) disappears, leaving you with just the original function without any surplus numerical value to consider.
For the problem \( \int -3 \csc^2 x \, dx \), adding \( C \) ensures that any vertical shift of the function \( -3 \cot x \) is represented. It signals that there are infinitely many functions which could be the antiderivative depending on the initial conditions. Always remember to include \( C \) for indefinite integrals to reflect this inherent flexibility.
This arises because the differentiation of a constant is zero. So, when checking your antiderivative by differentiation, the constant \( C \) disappears, leaving you with just the original function without any surplus numerical value to consider.
- Keeps solutions general rather than specific.
- Allows flexibility working with boundary conditions in definite integrals.
For the problem \( \int -3 \csc^2 x \, dx \), adding \( C \) ensures that any vertical shift of the function \( -3 \cot x \) is represented. It signals that there are infinitely many functions which could be the antiderivative depending on the initial conditions. Always remember to include \( C \) for indefinite integrals to reflect this inherent flexibility.
Other exercises in this chapter
Problem 47
Determine all critical points for each function. $$f(x)=x(4-x)^{3}$$
View solution Problem 47
Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\sqrt{|x|}=\left\\{\begin{array}{ll}\sqrt{-x},
View solution Problem 47
Give the acceleration \(a=d^{2} s / d t^{2},\) initial velocity. and initial position of an object moving on a coordinate line. Find the object's position at ti
View solution Problem 47
a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support
View solution