Problem 17 - University Calculus: Early Transcendentals

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. $$\text { a. } \csc x \cot x \quad \text { b. }-\csc 5 x \cot 5 x \quad \text { c. }-\pi \csc \frac{\pi x}{2} \cot \frac{\pi x}{2}$$

Step-by-Step Solution

Answer

a. \(-\csc x + C\), b. \(-\frac{1}{5}\csc(5x) + C\), c. \(-2\csc\left(\frac{\pi x}{2}\right) + C\)."

Step 1: Identify the Integral Formula

For part (a) \( \csc x \cot x \), recognize that the derivative of \( \csc x \) is \( -\csc x \cot x \). Therefore, an antiderivative of \( \csc x \cot x \) is \(-\csc x + C\), where \( C \) is the constant of integration.

Step 2: Differentiate to Check (a)

Differentiate \(-\csc x + C\) with respect to \( x \). The derivative is \(-(-\csc x \cot x) = \csc x \cot x\), which matches the original function. This confirms our antiderivative is correct.

Step 3: Apply the Transformation Rule (b)

For \( -\csc 5x \cot 5x \), recognize that this is connected to the derivative of \( \csc(5x) \). The antiderivative is thus \(-\frac{1}{5}\csc(5x) + C\) by using the chain rule and dividing by the inner derivative.

Step 4: Differentiate to Check (b)

Differentiate \(-\frac{1}{5}\csc(5x) + C\). The derivative is \(-\frac{1}{5}(-5\csc(5x)\cot(5x)) = -\csc(5x)\cot(5x)\), confirming the antiderivative matches the original function.

Step 5: Use Standard Results for Complex Expressions (c)

In \(-\pi \csc \left(\frac{\pi x}{2}\right) \cot \left(\frac{\pi x}{2}\right)\), recognize it relates to \(\csc\left(\frac{\pi x}{2}\right)\). Antiderivative is \(-2\csc\left(\frac{\pi x}{2}\right) + C\), multiplying by \( \frac{1}{\left(\frac{\pi}{2}\right)} \).

Step 6: Differentiate to Check (c)

Differentiate \(-2\csc\left(\frac{\pi x}{2}\right) + C\). The derivative is \(-2 \left(-\frac{\pi}{2}\csc\left(\frac{\pi x}{2}\right)\cot\left(\frac{\pi x}{2}\right)\right) = -\pi \csc\left(\frac{\pi x}{2}\right)\cot\left(\frac{\pi x}{2}\right)\), matching the original function.

Key Concepts

Integration Techniques

Integration is the process of finding antiderivatives and can involve various techniques. It's an essential mathematical method utilized in calculus to find the area under a curve. One commonly employed technique is recognizing derivatives of standard functions and inverting them to find the original function. The effectiveness of each technique often depends on the form and complexity of the integrand.

Some standard integration techniques include:

• Basic Power Rule: For a function of the form \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} + C \) if \( n eq -1 \).
• Substitution: Useful when the integrand is a composite function, and it simplifies the integral by making a substitution that simplifies the integrand.
• Integration by Parts: Applies to products of functions, following the formula \( \int u \, dv = uv - \int v \, du \).

For functions involving derivatives of basic trigonometric identities, like \( \csc x \cot x \), recognizing derivatives is crucial. Knowing that the derivative of \( \csc x \) is \(-\csc x \cot x\) leads directly to its antiderivative, demonstrating an efficient application of known derivative rules.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, appearing frequently in derivatives and antiderivatives. Understanding these functions and their derivatives is essential for efficiently solving integration problems. Specific trigonometric derivatives include:

• Derivative of \( \sin x \) is \( \cos x \).
• Derivative of \( \cos x \) is \(-\sin x \).
• Derivative of \( \tan x \) is \( \sec^2 x \).
• Derivative of \( \csc x \) is \(-\csc x \cot x \).
• Derivative of \( \sec x \) is \( \sec x \tan x \).
• Derivative of \( \cot x \) is \(-\csc^2 x \).

These derivatives form the basis for finding their corresponding antiderivatives. For instance, knowing the derivative of \( \csc x \) as \(-\csc x \cot x \) directly provides the antiderivative solution for certain integrands involving these functions. In this context, recognizing these standard derivatives simplifies the process of integration.

Chain Rule

The chain rule is a core concept in calculus that allows differentiation of composite functions. When studying integration, the chain rule is used in reverse to aid in finding antiderivatives, particularly when dealing with functions that have an inside function. This is crucial in cases where substitution might be necessary.

The chain rule operates as follows: If a function \( y \) depends on another function \( u \), which in turn depends on \( x \), and we are given \( y = f(u) \) and \( u = g(x) \), then the chain rule states:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]In terms of integration, if we recognize that a function was differentiated using the chain rule, its antiderivative can often be found by applying appropriate substitutions. For example, if one sees a term like \( -\csc(5x) \cot(5x) \), they might recognize it as a derivative of \( \csc(5x) \), thus the integral involves dividing by the derivative of the inside function, which is 5, resulting in \(-\frac{1}{5} \csc(5x) + C\). Such understanding allows both differentiation and integration to become more intuitive and accurate.