Problem 42 - Calculus for Biology and Medicine

Question

In Problems \(41-46\), assume that \(a\) is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\sin ^{2}\left(a^{2} x+1\right) $$

Step-by-Step Solution

Answer

The general antiderivative is \( F(x) = \frac{1}{2}x - \frac{1}{4a^2} \text{sin}(2a^2 x + 2) + C \).

Step 1: Understand the problem

We are asked to find the general antiderivative of the function \( f(x)= ext{sin}^2(a^2 x + 1) \). This means we need to find a function \( F(x) \) such that the derivative of \( F(x) \) gives us \( f(x) \).

Step 2: Utilize a trigonometric identity

To find an antiderivative, we'll use the identity \( ext{sin}^2( heta) = \frac{1 - ext{cos}(2\theta)}{2} \). Applying this identity to our function gives us \( ext{sin}^2(a^2 x + 1) = \frac{1 - ext{cos}(2(a^2 x + 1))}{2} \).

Step 3: Rewrite the function using the identity

The function becomes \( f(x) = \frac{1}{2} - \frac{1}{2} ext{cos}(2a^2 x + 2) \). This simplifies the process of finding an antiderivative.

Step 4: Find antiderivative of the first term

The antiderivative of the constant \( \frac{1}{2} \) is \( \frac{1}{2}x \), since the derivative of \( \frac{1}{2}x \) is \( \frac{1}{2} \).

Step 5: Find antiderivative of the second term

For \( -\frac{1}{2} ext{cos}(2a^2 x + 2) \), use the substitution \( u = 2a^2 x + 2 \), leading to \( du = 2a^2 dx \) or \( dx = \frac{du}{2a^2} \). Thus, the integral is \(-\frac{1}{2} \int ext{cos}(u) \frac{1}{2a^2} \ du = -\frac{1}{4a^2} \int ext{cos}(u) \, du \).

Step 6: Solve the resulting integral

The antiderivative of \( ext{cos}(u) \) is \( ext{sin}(u) \). So the integral \( -\frac{1}{4a^2} \int ext{cos}(u) \, du \) becomes \( -\frac{1}{4a^2} ext{sin}(u) \). Substitute back to terms of \( x \).

Step 7: Combine terms and include constant of integration

The general antiderivative is \( F(x) = \frac{1}{2}x - \frac{1}{4a^2} ext{sin}(2a^2 x + 2) + C \), where \( C \) is the constant of integration, representing the family of solutions.

Key Concepts

Trigonometric Identity

To solve problems involving antiderivatives of trigonometric functions, understanding and applying trigonometric identities is crucial. In the given function, we used the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \). This identity helps in transforming complex trigonometric expressions into simpler forms, making it easier to integrate or differentiate them.
This particular identity takes advantage of the angle doubling in the cosine function, effectively reducing the power in a sine function, which simplifies integration problems. By applying this identity to our original function \( f(x) = \sin^2(a^2 x + 1) \), we converted it to \( \frac{1}{2} - \frac{1}{2} \cos(2a^2 x + 2) \).

• The simplification using trigonometric identities can convert a difficult antiderivative into a sum or difference of simpler terms.
• Once simplified, integrating or finding derivatives becomes manageable due to familiarity with standard integral forms.

Recognizing these identities and practicing their application can significantly ease the problem-solving process.

Substitution Method

One of the most powerful techniques in calculus for finding antiderivatives is the substitution method. This involves substituting a part of the integrand with a new variable to simplify integration.
In our exercise, after simplifying \( f(x) \), we faced integrating \( -\frac{1}{2} \cos(2a^2 x + 2) \). We used substitution by letting \( u = 2a^2 x + 2 \). This substitution transforms the differential \( dx \) into \( \frac{du}{2a^2} \).
With this, the integral becomes simpler: \(-\frac{1}{2} \int \cos(u) \frac{1}{2a^2} \, du = -\frac{1}{4a^2} \int \cos(u) \, du \).

• Substitution often simplifies integration by reducing the integrand to a basic form or a look-up form.
• After integration is complete, always substitute back the original variable to express the solution in terms of \( x \).

Mastering this method allows you to tackle seemingly complex integrals with ease and is foundational in the study of calculus.

Integration Constant

Whenever you find an indefinite integral (antiderivative), you must always add the integration constant \( C \) to your solution. This constant accounts for all the possible values that the function could have taken before the derivative was applied, incorporating all vertical shifts of the antiderivative graph.
In this exercise, after computing the antiderivative as \( F(x) = \frac{1}{2}x - \frac{1}{4a^2} \sin(2a^2 x + 2) \), the term \( C \) is added to signify that this is the general solution.

• The constant \( C \) is crucial because it represents the infinity of solutions possible for an indefinite integral.
• In applications, \( C \) can be determined by initial or boundary conditions specific to the problem being solved.

Always remember to include this constant in your solutions, as omitting it implies that the antiderivative is uniquely defined when it is not. This key point ensures your solution is complete and mathematically correct.