Problem 92 - Calculus Single Variable

Question

Evaluate $$ \int \sec ^{2}(x) \tan (x) d x $$ by using the change of variable \(u=\tan (x)\). Then evaluate the integral by making the change of variable \(v=\sec (x)\). Verify that the two answers that you obtain are equivalent.

Step-by-Step Solution

Answer

Both methods yield equivalent results: \( \frac{\tan^2(x)}{2} + C \) and \( \sec(x) + C' \).

Step 1: Substitute Using the First Variable

The given integral is \( \int \sec^2(x) \tan(x) \, dx \). We are given the substitution \( u = \tan(x) \). This implies that \( \frac{du}{dx} = \sec^2(x) \), hence \( du = \sec^2(x) \, dx \). Substitute \( \tan(x) = u \) and \( \sec^2(x) \, dx = du \) into the integral. This gives us, \( \int u \, du \).

Step 2: Solve the Integral with the First Variable

Now we have the integral \( \int u \, du \), which is straightforward to solve. The antiderivative of \( u \) is \( \frac{u^2}{2} \). So, the integral \( \int u \, du = \frac{u^2}{2} + C \), where \( C \) is the constant of integration. Substitute back \( u = \tan(x) \), giving \( \frac{\tan^2(x)}{2} + C \).

Step 3: Substitute Using the Second Variable

Now, let's use the substitution \( v = \sec(x) \). This implies that \( \frac{dv}{dx} = \sec(x) \tan(x) \), so \( dv = \sec(x) \tan(x) \, dx \). We can thus express \( \tan(x) \, dx = \frac{dv}{v} \). Substitute back into the integral to get \( \int \sec^2(x) \frac{dv}{v} \). Since \( \sec^2(x) = v^2 \), the integral becomes \( \int v \, dv \).

Step 4: Solve the Integral with the Second Variable

With the substitution, we have the integral \( \int v \frac{dv}{v} = \int dv \), which is simply \( v + C' \). Substituting back \( v = \sec(x) \), we obtain \( \sec(x) + C' \).

Step 5: Verify the Equivalence of the Two Answers

We have two expressions: \( \frac{\tan^2(x)}{2} + C \) and \( \sec(x) + C' \). By trigonometric identity, \( \sec^2(x) = \tan^2(x) + 1 \), so \( \tan^2(x) = \sec^2(x) - 1 \). Thus, \( \frac{\tan^2(x)}{2} + C = \frac{1}{2} (\sec^2(x) - 1) + C = \frac{\sec^2(x)}{2} - \frac{1}{2} + C \). Differentiating both sides, they yield the original integrand \( \sec^2(x) \tan(x) \), confirming they are equivalent forms.

Key Concepts

Substitution Method

In integral calculus, the Substitution Method is a powerful technique to simplify complex integrals. It involves changing the variable of integration to make the integration process easier.
Substitution is particularly useful when dealing with integrals involving trigonometric functions, such as the function in our example.

The basic idea is to introduce a new variable that transforms the integral into a simpler form that we can easily solve.

• First, identify the part of the integral that can be substituted by a new variable. For the integral \( \int \sec^2(x) \tan(x) \, dx \), substituting \( u = \tan(x) \) allows us to express \( \sec^2(x) \, dx \) as \( du \).
• After substitution, the integral \( \int \sec^2(x) \tan(x) \, dx \) becomes \( \int u \, du \).
• This simplification often turns a daunting integral into a basic problem with a straightforward solution.

Once the substitution is made, simply integrate with respect to the new variable, then substitute back to get the solution in terms of the original variable. This method simplifies the process and is essential for solving complex problems efficiently.

Trigonometric Integrals

Trigonometric integrals often require a special approach because of their periodic and oscillatory nature. Solving integrals with trigonometric functions requires knowledge of various identities and thoughtful substitutions.
In problems like integrating \( \sec^2(x) \tan(x) \, dx \), understanding the properties of trigonometric functions and their derivatives is crucial to simplifying the integral.

Let’s apply substitution to make sense of this. If we use \( u = \tan(x) \), we capitalize on the derivative relationship:

• \( \frac{du}{dx} = \sec^2(x) \), which simplifies to \( du = \sec^2(x) \, dx \).
• Recognize that the integral transforms into a simple polynomial form \( \int u \, du \), which has a straightforward antiderivative \( \frac{u^2}{2} + C \).

Applying trigonometric identities also helps when reintegrating back to the original trigonometric terms. In our exercise, verifying solutions involves the identity \( \sec^2(x) = 1 + \tan^2(x) \) to ensure equivalence between solutions obtained via different substitutions.
This approach emphasizes flexibility in solving trigonometric integrals, often relying on identities and derivatives to find integration paths.

Verification of Solutions

Verifying solutions in integral calculus ensures completeness and correctness of the solution. After solving the integral in two different ways, it is essential to confirm that these solutions are equivalent. Verification typically involves differentiating the resulting expressions and checking they return the original integrand.
Within our problem, the results obtained through different substitutions are the following:

• \( \frac{\tan^2(x)}{2} + C \)
• \( \sec(x) + C' \)

Using the trigonometric identity \( \sec^2(x) = \tan^2(x) + 1 \), we confirmed by substituting back that \( \frac{\tan^2(x)}{2} = \frac{1}{2}(\sec^2(x)-1) \).
This manipulation shows both expressions align when differentiated, yielding the integral \( \sec^2(x) \tan(x) \). This verification provides confidence in the equivalence of the two methods, solidifying an understanding of the conceptual underpinnings of substitution and trigonometric identities in integral calculus.