Problem 21 - Biocalculus Calculus for the Life Sciences

Question

First make a substitution and then use integration by parts to evaluate the integral. \(\int \cos \sqrt{x} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(2(\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}) + C.\)

Step 1: Substitute to Simplify the Integral

Let us substitute by setting \( u = \sqrt{x} \), which implies \( x = u^2 \). Then, the differential \( dx \) becomes \( 2u \, du \). Therefore, the integral \( \int \cos \sqrt{x} \, dx \) becomes \( \int \cos u (2u) \, du = 2 \int u \cos u \, du \).

Step 2: Apply Integration by Parts Formula

Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \). Here, set \( v = \sin u \) and \( dv = \cos u \, du \). Choose \( u = u \), which gives \( du = 1 \, du \).

Step 3: Integrate \( \int u \cos u \, du \)

Using integration by parts, \( \int u \cos u \, du = u \sin u - \int \sin u \, du \).Evaluate the second integral: \( \int \sin u \, du = -\cos u \).Thus, \( u \sin u - (-\cos u) = u \sin u + \cos u \).

Step 4: Substitute Back to Original Variable

Replace \( u \) back with \( \sqrt{x} \) as \( u = \sqrt{x} \). So the integral becomes \( 2(\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}) + C \) where \( C \) is the integration constant.

Key Concepts

Substitution Method

The Substitution Method in calculus is a powerful tool for simplifying integrals, especially when you encounter complex functions within the integral that make direct integration challenging. This method is akin to using a change of variables in order to transform the integral into a simpler form. This is much like unraveling a complex puzzle into smaller, manageable pieces.

To effectively use substitution:

• Identify the part of the integral to substitute. Often, this is under a radical or inside a trigonometric function.
• Set a new variable, say \( u \), to equal this part. In our problem, we set \( u = \sqrt{x} \).
• Determine \( dx \) in terms of \( du \). This involves differentiating the expression. Here, since \( x = u^2 \), we derived \( dx = 2u \, du \).
• Rewrite the entire integral in terms of \( u \). This simplification makes the integration process more manageable.

Once the substitution is complete, the problem often transforms into a standard integral that is easier to evaluate. Don't forget to substitute back your original variable at the end.

Integration by Parts

Integration by Parts is another fundamental technique in calculus, relying on the product rule for differentiation. It is particularly useful when you need to integrate the product of two functions. This technique helps in breaking down complex integrals into simpler ones.

The formula is expressed as \( \int u \, dv = uv - \int v \, du \). Here's how it works:

• Select parts of the integrand to designate as \( u \) and \( dv \). Here, choosing \( u = u \) and \( dv = \cos u \, du \) was strategic.
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \). In this case, \( du = 1 \, du \) and \( v = \sin u \).
• Substitute into the formula to simplify: \( \int u \cos u \, du = u \sin u - \int \sin u \, du \).

Continuing the integration works by incrementally reducing the complexity of the problem, which ultimately leads to a solution that combines these parts into an integrated whole.

Definite and Indefinite Integrals

In calculus, understanding the distinction between definite and indefinite integrals is crucial for solving problems. They serve different purposes in mathematics, and recognizing when to use each is key.

Indefinite Integrals

• Represent a family of functions and include an arbitrary constant \( C \).
• Indefinite integrals are essentially antiderivatives of the function, generalizing the notion of accumulation under the curve.
• The solution in our example, \( 2(\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}) + C \), is an indefinite integral, reflecting a general solution.

By contrast,

Definite Integrals

• Calculate the exact area under the curve between two points. They have limits of integration.
• The output is a specific numerical value, not an expression with an integration constant.

Each type of integral plays a unique role in solving conceptual and real-world problems, understanding this difference allows for apt application in various calculus challenges.