Problem 60 - Thomas Calculus in SI Units

Question

Verify the formulas by differentiation. $$\int \csc ^{2}\left(\frac{x-1}{3}\right) d x=-3 \cot \left(\frac{x-1}{3}\right)+C$$

Step-by-Step Solution

Answer

Differentiating yields the integrand, so the formula is verified.

Step 1: Understand the Formula to Verify

The formula given is \( \int \csc^{2}\left(\frac{x-1}{3}\right) dx =-3 \cot\left(\frac{x-1}{3}\right) + C \). To verify, we need to differentiate the right side and check that it equals the integrand, \( \csc^{2}\left(\frac{x-1}{3}\right) \).

Step 2: Differentiate the Right Side

Differentiate \(-3 \cot\left(\frac{x-1}{3}\right) + C \) with respect to \( x \). Use the chain rule: the derivative of \( \cot(u) \) is \(-\csc^{2}(u)\), and the derivative of \( \frac{x-1}{3} \) with respect to \( x \) is \( \frac{1}{3} \). Therefore, the differentiation process gives \(-3 \times \left(-\csc^{2}\left(\frac{x-1}{3}\right)\right) \times \frac{1}{3} \).

Step 3: Simplify the Differentiated Expression

Simplify the differentiated expression: \(-3 \times -\csc^{2}\left(\frac{x-1}{3}\right) \times \frac{1}{3} = \csc^{2}\left(\frac{x-1}{3}\right)\).

Step 4: Verify the Result

The simplified derivative \( \csc^{2}\left(\frac{x-1}{3}\right) \) matches the original integrand. Therefore, the initial formula is verified to be correct.

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with the differentiation of composite functions. A composite function is when one function is placed inside another, like a nested function. In our exercise, the function is composed of \( \cot \left( \frac{x-1}{3} \right) \). To differentiate this, the chain rule plays a key role.

When applying the chain rule, first differentiate the outer function, then multiply by the derivative of the inner function. Here’s a quick reminder:

• Outer function: \( \cot(u) \)
• Inner function: \( u = \frac{x-1}{3} \)

The derivative of \( \cot(u) \) is \(-\csc^2(u)\). Differentiating the inner function, \( \frac{x-1}{3} \), gives us \( \frac{1}{3} \). Therefore, applying the chain rule means multiplying these results together. This calculation ensures you properly address each layer of the function.

Trigonometric Integration

Trigonometric integration involves the antiderivatives of trigonometric functions, like \( \csc^2(u) \), \( \sin(u) \), or \( \cos(u) \). When you integrate \( \csc^2(u) \), it uniquely results in \(-\cot(u)\), usually plus a constant of integration, \( C \).

This relationship is significant in calculus. Knowing these integral-identities by heart streamlines both solving problems and verifying results. In our exercise, you started with the integral of \( \csc^2\left(\frac{x-1}{3}\right) \) and related it to \(-3 \cot\left(\frac{x-1}{3}\right) + C\).

• The process clarifies how integration and differentiation are inverse operations, particularly clear in this trigonometric context.
• The integral forms provide shortcuts in integrating complex functions once their trigonometric identity is recognized.

Given the often cyclical nature of trigonometric functions, their integrals and derivatives can end up being quite intuitive!

Differentiation

Differentiation is a core pillar in calculus, used to find the rate at which a function is changing at any given point. In this exercise, you verified the formula by differentiation.

Start by recognizing that you're differentiating the result of integration to check your work. This is a classic approach to ensure your integral was correctly executed. Here, you differentiated \(-3 \cot\left(\frac{x-1}{3}\right) + C \) and verified it matched the integrand \( \csc^2\left( \frac{x-1}{3} \right) \).

• Ensure seamless application of differentiation rules, especially the chain rule when confronting composite functions.
• Notice how constants like \( C \) disappear during differentiation, as constants do not affect the rate of change.

Anyone wanting to master calculus should regularly practice these kinds of problems to reinforce a balanced understanding of both integration and differentiation practices.