Problem 29 - Calculus for Biology and Medicine

Question

Use substitution to evaluate the indefinite integrals. $$ \int \sin \left(\frac{3 \pi}{2} x+\frac{\pi}{4}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{2}{3\pi} \cos\left(\frac{3\pi}{2} x + \frac{\pi}{4}\right) + C\)."

Step 1: Identify the substitution

Let's set the substitution as follows: \( u = \frac{3\pi}{2} x + \frac{\pi}{4} \). This substitution simplifies the sine function.

Step 2: Differentiate the substitution

Differentiate \( u \) with respect to \( x \) to find \( du \): \[ \frac{du}{dx} = \frac{3\pi}{2} \] Thus, \( du = \frac{3\pi}{2} dx \).

Step 3: Solve for dx

Rearrange the expression from the previous step to solve for \( dx \):\[ dx = \frac{2}{3\pi} du \]

Step 4: Substitute in the integral

Substitute \( u \) and \( dx \) into the original integral:\[ \int \sin\left(\frac{3\pi}{2} x + \frac{\pi}{4}\right) dx = \int \sin(u) \left(\frac{2}{3\pi} du \right) \]This simplifies to:\[ \frac{2}{3\pi} \int \sin(u) \, du \]

Step 5: Integrate with respect to u

Integrating \( \int \sin(u) \, du \) gives:\[ \int \sin(u) \, du = -\cos(u) + C \] Therefore, the integral becomes:\[ \frac{2}{3\pi} \left(-\cos(u)\right) + C = -\frac{2}{3\pi} \cos(u) + C \]

Step 6: Substitute back for u

Replace \( u \) with its original expression:\[ -\frac{2}{3\pi} \cos\left(\frac{3\pi}{2} x + \frac{\pi}{4}\right) + C \] This is the evaluation of the indefinite integral in terms of \( x \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus, designed to simplify the integration process by transforming a complicated integral into an easier form. It is akin to the chain rule in differentiation, helping us to work with composite functions more easily.

When you identify a part of an integral that can be replaced with a single variable, like "\( u \)," you're using substitution. This process involves a few clear steps:

• Identify part of the integrand to set as "\( u \)," often by choosing a function whose derivative is also present in the integrand.
• Differential transformation: Differentiate "\( u \)" with respect to "\( x \)" and solve for "\( dx \)" in terms of "\( du \)." This is essential to swap the variables.
• Substitute: Replace the identified part of the integral (\( \sin(\frac{3\pi}{2} x + \frac{\pi}{4}) \)) and "\( dx \)" with "\( u \)" and the equivalent "\( du \)" expression.
• Integrate the new simplified function often expressed in terms of "\( u \)."

This method greatly simplifies the integration of more complex, composite, or trigonometric functions.

Integral Calculus

Integral calculus, the counterpart to differential calculus, focuses on accumulation—finding the overall impact given a rate of change. It involves finding functions known as antiderivatives or integrals. One common type is the indefinite integral which doesn't have specified limits, hence includes an arbitrary constant denoted as "\( C \)."

When performing integral calculus, particularly indefinite integrals, several methods exist:

• Basic antiderivatives, where known forms are integrated directly.
• Substitution method, which offers great utility when faced with functions that are derivatives of other functions present in the integrand.
• Integration by parts, useful for product functions, though not applied in the current problem.

The indefinite integral is represented as \( \int f(x) \, dx = F(x) + C \), where "\( F(x) \)" is an antiderivative of the function "\( f \)."

In the given exercise, we encounter the integration of trigonometric functions such as \( \sin(u) \), recognizing that indefinite integrals help revert the change described by derivatives, thereby revealing the original function that yielded the derivative.

Trigonometric Integrals

Trigonometric integrals are those that involve trigonometric functions such as sine, cosine, tangent, and their inverses. These functions are often periodic and require specific strategies for integration.

In the context of trigonometric integrals, substitution is often employed to handle them effectively. For example, recognizing forms like \( \int \sin(u) \, du \), simplifies to \( -\cos(u) + C \). This direct integration of trigonometric functions highlights the anti-derivatives of basic trigonometric functions:

• \( \int \sin(u) \, du = -\cos(u) + C \)
• \( \int \cos(u) \, du = \sin(u) + C \)

In our specific exercise, integrating \( \sin(u) \) with respect to "\( u \)" provides this easier path allowing transformation back into terms of the original variable, typically yielding an expression with the cosine of the initial trigonometric argument.

Such integrals are common in physics, engineering, and other fields where waveforms or periodic phenomena are modeled.