Problem 64 - Thomas Calculus in SI Units

Question

Right, or wrong? Say which for each formula and give a brief reason for each answer. a. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{\sec ^{3} \theta}{3}+C\) b. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{1}{2} \tan ^{2} \theta+C\) c. \(\int \tan \theta \sec ^{2} \theta d \theta=\frac{1}{2} \sec ^{2} \theta+C\)

Step-by-Step Solution

Answer

a: Wrong, b: Right, c: Wrong

Step 1: Understand the Integral

We need to evaluate the integral \( \int \tan \theta \sec^2 \theta \, d\theta \). The integrand can be simplified using trigonometric identities, where \( \sec^2 \theta = 1 + \tan^2 \theta \), which suggests trying integration by substitution.

Step 2: Perform Substitution for the Integral

Let \( u = \tan \theta \), then \( du = \sec^2 \theta \, d\theta \). Substitute \( u \) to get the integral \( \int u \, du \), which simplifies to \( \frac{1}{2} u^2 + C \).

Step 3: Back Substitute \( u \)

Since \( u = \tan \theta \), replacing \( u \) back we have the antiderivative \( \frac{1}{2} \tan^2 \theta + C \).

Step 4: Validate Formula a

The formula a states \( \int \tan \theta \sec^2 \theta \, d\theta = \frac{\sec^3 \theta}{3} + C \). Our calculated antiderivative is \( \frac{1}{2} \tan^2 \theta + C \), which does not match formula a. Hence, formula a is **wrong**.

Step 5: Validate Formula b

The formula b states \( \int \tan \theta \sec^2 \theta \, d\theta = \frac{1}{2} \tan^2 \theta + C \). This formula matches exactly with our derived antiderivative, making formula b **right**.

Step 6: Validate Formula c

The formula c states \( \int \tan \theta \sec^2 \theta \, d\theta = \frac{1}{2} \sec^2 \theta + C \). Our derived antiderivative is \( \frac{1}{2} \tan^2 \theta + C \), which differs from the expression given in formula c. Therefore, formula c is **wrong**.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used in calculus to simplify difficult integrals. It works like "reversing" the chain rule from differentiation.

The idea is to change the variable of integration to make the integral simpler. For example, in many integrals involving trigonometric functions, substitution can make the calculation much easier.

For the integral \( \int \tan \theta \sec^2 \theta \, d\theta \), the substitution \( u = \tan \theta \) simplifies the integration process. We then differentiate \( u \) to get \( du = \sec^2 \theta \, d\theta \), making the integral \( \int u \, du \). This technique effectively changes a complex integral into a simpler one.

• The act of substitution helps address the composition of functions in the integrand.
• It converts the problem into an integral related to the simpler function \( u \).

Once integration is performed, we substitute back the original variable to obtain the solution in the proper context.

Remember, substitution is particularly useful when you can identify a function within the integrand whose derivative is also present.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions and can often be solved using identities and substitution.

These integrals frequently appear in calculus, and using trigonometric identities can simplify them significantly. For example, the identity \( \sec^2 \theta = 1 + \tan^2 \theta \) can be useful in solving these types of problems.

In our exercise, recognizing \( \tan \theta \sec^2 \theta \) can be related through its derivative \( \sec^2 \theta \/d\theta = \tan \theta \), which helps simplify the integration process.

• Understanding identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) can help rotate the forms and simplify them for easier integration.
• Additionally, patterns such as the derivatives of trigonometric functions guide valid substitutions.

Ultimately, manipulating these functions via identities and strategic substitutions can greatly simplify trigonometric integrals.

Integral Calculus Concepts

Integral calculus focuses on the concept of determining the area under curves, and it plays a crucial role in understanding the accumulation of quantities.

The fundamental theorem of calculus bridges the concept of derivative as the rate of change and integral as total accumulation. Integrals can be definite, giving a numerical result, or indefinite, providing a family of functions represented by the function plus a constant \( C \).

• Indefinite integrals result in antiderivatives, which are general representations inclusive of a constant \( C \).
• Definite integrals compute exact values by setting boundaries on the integral.

In the context of solving integrals like \( \int \tan \theta \sec^2 \theta \, d\theta \), it is crucial to explore and apply different techniques—such as substitution—to arrive at the correct antiderivative, thereby unraveling the problem's solution through careful calculation and reasoning.

Overall, integral calculus provides the tools necessary to calculate problems related to areas, volumes, and other quantities that accumulate continuously.