Question

In Exercises \(7-12,\) use differentiation to verify the antiderivative formula. $$\int \csc u \cot u=-\csc u+C$$

Short answer

The differentiation of \(-\csc(u) + C\) gives \(\csc(u) \cdot \cot(u)\), verifying that \(-\csc(u) + C\) is indeed the antiderivative of \(\csc(u) \cdot \cot(u)\).

Step-by-step solution

Write down the formula

We wish to verify that \(-\csc u + C\) is indeed the antiderivative of \(\csc u \cot u\), which means that the derivative of \(-\csc u + C\) should yield \(\csc u \cot u\).

Differentiate the function \(-\csc u + C\)

Differentiation of \(-\csc u + C\) should yield \(\csc u \cdot \cot u\). Since the derivative of a constant C is always 0, we only need to find the derivative of \(-\csc u\). The derivative of \(-\csc u\) can be found using the rule that the derivative of \(csc u\) is \(-csc u \cdot cot u\). As such the derivative of \(-\csc u\), is \(-(-\csc u \cdot cot u) = \csc u \cdot cot u\)

Compare the results

Comparing the result of the differentiation with the original integrand, they both match. Thus, it can be verified that \(-\csc u + C\) is indeed the antiderivative of \(\csc u \cdot \cot u\). If you differentiate \(-\csc u + C\), you get \(\csc u \cdot \cot u\).

Key concepts

Differentiation

Understanding the concept of differentiation is imperative for verifying the antiderivative of a function. Differentiation is a process in calculus which determines the rate at which a function is changing at any given point. It's like taking a snapshot of the function at a particular instant, and seeing how steep the curve is at that point, which tells us how fast the function's value is going up or down.

Let's simplify this with an example. Imagine you're on a hike, and you're tracking the elevation change as you walk. If you could measure exactly how steep the hike is at every step, you'd be doing something similar to differentiation. Now, in terms of our exercise, we are given the function \( -\text{csc}(u) \). Our goal is to find its derivative, which will tell us the steepness or rate of change of \( -\text{csc}(u) \) at any point \(u\).

The technique for finding the derivative involves applying rules that have been established for different types of functions. In the case of trigonometric functions like cosecant, we have specific rules that help us calculate derivatives without having to create the limit definition of the derivative every single time, which would be time-consuming and cumbersome. By mastering the art of differentiation, one not only verifies antiderivatives but also unlocks the ability to analyze the behavior of functions dynamically.

Cosecant Function

In connection with our problem, let's delve into the cosecant function, often denoted as \( \text{csc}(u) \). It is one of the six fundamental trigonometric functions and is the reciprocal of the sine function. This means that for any angle \(u\), \( \text{csc}(u) = \frac{1}{\text{sin}(u)} \). While less common than sine or cosine in introductory courses, cosecant plays an important role in various calculus operations, including differentiation and integration.

In practical terms, if you were to look at the unit circle, which is a circle with a radius of 1 centered at the origin of a coordinate system, the value of \( \text{csc}(u) \) would be the length of the line segment from the origin to the point where the extension of the line from the circle, through the point at angle \(u\), reaches the vertical tangent to the circle. It's a bit of a mouthful, but visualizing this can make understanding trigonometric functions much clearer.

As part of our primary exercise, we need to grasp that the derivative of \( \text{csc}(u) \) is \( -\text{csc}(u) \text{cot}(u) \). This is a critical step in verifying the antiderivative by differentiation. Remembering the derivatives of trigonometric functions is vital as it makes solving calculus problems, especially integration and differentiation, more manageable and less time-consuming.

Cotangent Function

Our exercise also involves the cotangent function, denoted as \( \text{cot}(u) \). Similar to cosecant, cotangent is the reciprocal of another well-known function, the tangent. To be precise, \( \text{cot}(u) = \frac{1}{\text{tan}(u)} \) or equivalently, \( \text{cot}(u) = \frac{\text{cos}(u)}{\text{sin}(u)} \). In the context of a circle, cotangent can represent the ratio of the adjacent side to the opposite side of a triangle inscribed in the unit circle where the angle at the center is \(u\).

For the calc-savvy, understanding how trigonometric functions like cotangent behave, and how they change - which is where differentiation comes in - is essential. When introduced in conjunction with the cosecant function, as we see in the expression \( \text{csc}(u) \text{cot}(u) \), it can initially seem daunting to differentiate. However, once you remember the unique properties and derivatives of trigonometric functions, the process becomes far less intimidating.

Understanding the cotangent function not only aids in learning to differentiate but is also a key player in solving many practical problems in fields involving waves, resonance, and even in the construction of computer algorithms. Grasping both cotangent and cosecant, along with their relation to their better-known counterparts, sine and tangent, builds a strong foundation for mastering calculus concepts.