Problem 42 - Calculus for Biology and Medicine

Question

Assume that a is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\sin ^{2}(a x+1) $$

Step-by-Step Solution

Answer

The general antiderivative is \( \frac{1}{2}ax + \frac{1}{2} - \frac{1}{4}\sin(2ax + 2) + C \).

Step 1: Identify the Inner Function

The given function is \( f(x) = \sin^2(ax + 1) \). The inner function is \( u = ax + 1 \). Differentiating this, we get \( \frac{du}{dx} = a \).

Step 2: Express Function in Terms of u

Using the substitution \( u = ax + 1 \), we rewrite the function as \( \sin^2(u) \). Now, we'll find the antiderivative of \( \sin^2(u) \).

Step 3: Use Trigonometric Identity

Use the trigonometric identity \( \sin^2(u) = \frac{1}{2}(1 - \cos(2u)) \). Thus, we can rewrite the function as \( \frac{1}{2}(1 - \cos(2(ax + 1))) \).

Step 4: Antiderivate Each Term

Now we will find the antiderivative of \( \frac{1}{2}(1 - \cos(2(ax + 1))) \). Consider: \( \int \frac{1}{2}(1 - \cos(2u)) \, du \). Separate the integral: \( \frac{1}{2} \int 1 \, du - \frac{1}{2} \int \cos(2u) \, du \).

Step 5: Integrate Each Component

For \( \int 1 \, du \), the antiderivative is \( u \). For \( \int \cos(2u) \, du \), use substitution \( v = 2u \), \( dv = 2 \, du \), so \( du = \frac{dv}{2} \). The integral becomes: \( \frac{1}{2} \int \cos(v) \frac{dv}{2} = \frac{1}{4} \sin(v) = \frac{1}{4} \sin(2u) \).

Step 6: Combine and Substitute Back

The antiderivative becomes: \( \frac{1}{2}u - \frac{1}{4}\sin(2u) + C \). Substituting back \( u = ax + 1 \), we have: \( \frac{1}{2}(ax + 1) - \frac{1}{4}\sin(2(ax + 1)) + C \).

Step 7: Simplify Final Expression

Simplify the expression for clarity: \( \frac{1}{2}ax + \frac{1}{2} - \frac{1}{4}\sin(2ax + 2) + C \). This is the general antiderivative.

Key Concepts

Antiderivatives

In calculus, antiderivatives represent a core concept used to reverse the process of differentiation. An antiderivative of a function is basically a function that, when differentiated, will give you the original function back. If you have a function \( f(x) \), its antiderivative is usually denoted as \( F(x) \) and can be expressed as \( \int f(x) \, dx \). Because antiderivatives include an arbitrary constant \( C \), this constant accounts for any constant term that differentiation wipes away. So the general form of an antiderivative of \( f(x) \) is \( F(x) + C \).
This is important for solving area problems, among other applications. When you calculate the integral to find an antiderivative, it's like searching for the function that diesn't just dwell beneath the given function but embraces it, simplifying and shaping it into something useable. This makes antiderivatives essential for calculus and mathematical analysis.

Trigonometric Identities

Trigonometric identities are critical for simplifying complex trigonometric expressions within calculus. These identities represent equations involving trig functions that hold true for all values within their domains. In the context of the given problem, we use the identity \( \sin^2(u) = \frac{1}{2}(1 - \cos(2u)) \).

• This particular identity helps transform products or squares of sine into a linear cosine term, significantly simplifying integration.
• They enable us to express trigonometric functions in different ways, which can also be advantageous when solving integrals or derivatives.

Understanding trigonometric identities can drastically simplify solving calculus exercises that involve trigonometry, making them easier to integrate or differentiate.

Substitution Method

The substitution method is a powerful technique in calculus for simplifying the calculation of integrals. It involves changing variables to simplify an integral into a form that is easier to solve. By introducing a new variable, say \( u \), into an integral and expressing the integral in terms of \( u \), the integration becomes more straightforward.

• For the function \( \sin^2(ax + 1) \), we recognize \( ax + 1 \) as a suitable inner function \( u \) which simplifies some expressions.
• By substituting \( u = ax + 1 \), and evaluating the differential \( du = a \, dx \), the wider integral transforms into terms of \( u \), facilitating easier computation.
• After finding the integral in terms of \( u \), substituting back to the original variable ensures the result fits the original problem context.

The substitution method is essential for tackling complex integrals, allowing calculus problems to become manageable and solvable.