Problem 33
Question
In Problems 1-40, find the general antiderivative of the given function. $$ f(x)=\sec ^{2}\left(\frac{x}{3}\right) $$
Step-by-Step Solution
Verified Answer
The general antiderivative is \( F(x) = 3 \tan \left( \frac{x}{3} \right) + C \).
1Step 1: Identify the Function
The given function is \[ f(x) = \sec^2 \left( \frac{x}{3} \right) \] This resembles the derivative of the tangent function.
2Step 2: Recall the Antiderivative Rule
Recall that the derivative of \( \tan(x) \) is \( \sec^2(x) \). Therefore, the antiderivative of \( \sec^2(x) \) is \( \tan(x) \).
3Step 3: Perform Adjustments for Inner Function
Notice the inner function \( \frac{x}{3} \). To adjust for this, use substitution: let \( u = \frac{x}{3} \), so \( du = \frac{1}{3}dx \) or \( dx = 3du \).
4Step 4: Integrate Using Substitution
Now substitute: \[ \int \sec^2 \left( \frac{x}{3} \right) dx \] can be transformed to \[ 3 \int \sec^2(u) du \]The antiderivative of \( \sec^2(u) \) is \( \tan(u) \), so it becomes \[ 3 \tan(u) + C \]
5Step 5: Substitute Back for x
Substitute \( u = \frac{x}{3} \) back into the expression: \[ 3 \tan \left( \frac{x}{3} \right) + C \] Thus, the general antiderivative is \[ F(x) = 3 \tan \left( \frac{x}{3} \right) + C \]
6Step 6: Final Step: Write the General Solution
The general antiderivative or integral of the function is: \[ F(x) = 3 \tan \left( \frac{x}{3} \right) + C \] where \( C \) is an arbitrary constant of integration.
Key Concepts
IntegrationSubstitution MethodTrigonometric Functions
Integration
Integration is a fundamental concept in calculus that serves as the reverse process of differentiation. It involves finding the antiderivative or the original function when its derivative is known. Integration is used across various fields such as physics, engineering, and economics to calculate areas under curves, volumes of solids of revolution, and to solve differential equations.
In mathematics, we denote the integration operation by the symbol \( \int \). The purpose of integration is to accumulate quantities, which might be area, volume, or other values, depending on the type of function used.
In the exercise provided, the goal is to find the general antiderivative of the function \( f(x) = \sec^2 \left( \frac{x}{3} \right) \). This involves determining a function whose derivative is the given function. The antiderivative is essential as it represents a family of functions that differ only by a constant \( C \), known as the constant of integration.
To sum up, integration helps us "piece together" a function from its rate of change. Remember, the main task in antiderivatives is reversing the action that derivatives perform.
In mathematics, we denote the integration operation by the symbol \( \int \). The purpose of integration is to accumulate quantities, which might be area, volume, or other values, depending on the type of function used.
In the exercise provided, the goal is to find the general antiderivative of the function \( f(x) = \sec^2 \left( \frac{x}{3} \right) \). This involves determining a function whose derivative is the given function. The antiderivative is essential as it represents a family of functions that differ only by a constant \( C \), known as the constant of integration.
To sum up, integration helps us "piece together" a function from its rate of change. Remember, the main task in antiderivatives is reversing the action that derivatives perform.
Substitution Method
The substitution method, often nicknamed "u-substitution," is a technique used to simplify complex integrals. It involves replacing a part of the integral with a single variable, which makes the integration process more straightforward.
For instance, in our problem with \( \int \sec^2 \left( \frac{x}{3} \right) dx \), we identify the inner function \( \frac{x}{3} \). To handle this effectively, we let \( u = \frac{x}{3} \). Consequently, we compute the corresponding differential \( du = \frac{1}{3} dx \) or rearranged to \( dx = 3 du \).
By substituting \( u \) and \( dx \), the integral simplifies to \( 3 \int \sec^2(u) du \). This transformation is strategic as \( \int \sec^2(u) \) is more recognizable and straightforward to integrate.
Overall, the substitution method is invaluable for transforming complicated expressions into simpler forms, easing the integration process substantially.
For instance, in our problem with \( \int \sec^2 \left( \frac{x}{3} \right) dx \), we identify the inner function \( \frac{x}{3} \). To handle this effectively, we let \( u = \frac{x}{3} \). Consequently, we compute the corresponding differential \( du = \frac{1}{3} dx \) or rearranged to \( dx = 3 du \).
By substituting \( u \) and \( dx \), the integral simplifies to \( 3 \int \sec^2(u) du \). This transformation is strategic as \( \int \sec^2(u) \) is more recognizable and straightforward to integrate.
Overall, the substitution method is invaluable for transforming complicated expressions into simpler forms, easing the integration process substantially.
Trigonometric Functions
Trigonometric functions, like sine, cosine, and tangent, are pervasive in calculus. These functions form the backbone of many integration and differentiation problems and have specific rules when it comes to their derivatives and antiderivatives.
In our exercise, the function involved is \( \sec^2(x) \), which is related to the derivative of the tangent function: \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \).
Understanding trigonometric identities and derivatives is crucial when performing calculus operations. The antiderivative of \( \sec^2(x) \) is \( \tan(x) \), a fundamental relationship that was leveraged in solving the problem at hand.
In our exercise, the function involved is \( \sec^2(x) \), which is related to the derivative of the tangent function: \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \).
Understanding trigonometric identities and derivatives is crucial when performing calculus operations. The antiderivative of \( \sec^2(x) \) is \( \tan(x) \), a fundamental relationship that was leveraged in solving the problem at hand.
- \( \int \sec^2(x) dx = \tan(x) + C \)
- Trigonometric integral problems often involve recognizing these relationships to find solutions efficiently.
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