Problem 53 - Calculus for Biology and Medicine

Question

In Problems 49-60, use either substitution or integration by parts to evaluate each integral. $$ \int 2 x \sin \left(x^{2}\right) d x $$

Step-by-Step Solution

Answer

The integral \( \int 2x \sin(x^2) \, dx = -\cos(x^2) + C \).

Step 1: Choose the Integration Method

For the integral \( \int 2x \sin(x^2) \ dx \), substitution method is appropriate because the derivative of \( x^2 \) is \( 2x \), which appears in the integral.

Step 2: Perform Substitution

Let \( u = x^2 \), then \( du = 2x \, dx \). This changes the integral to \( \int \sin(u) \, du \).

Step 3: Integrate with Respect to u

The integral \( \int \sin(u) \, du \) is evaluated as \( -\cos(u) + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back to x

Replace \( u \) with \( x^2 \) to get the final result: \( -\cos(x^2) + C \).

Step 5: State the Final Result

The evaluated integral is \( -\cos(x^2) + C \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus integration. It is particularly useful when the integral contains a function and its derivative. In the context of the integral \( \int 2x \sin(x^2) \, dx \), substitution helps simplify the integration process.

To effectively use this method:

• Identify parts of the integrand that resemble a derivative. In this case, \( x^2 \), whose derivative is \( 2x \), is a good candidate.
• Set a substitution such that \( u = x^2 \) and consequently, \( du = 2x \, dx \). This substitution transforms the integral into \( \int \sin(u) \, du \), simplifying the integration task.
• Integrate with respect to the new variable \( u \). The integral \( \int \sin(u) \, du \) results in \( -\cos(u) + C \).
• Finally, replace back the original variable to get the solution in terms of \( x \), resulting in \( -\cos(x^2) + C \).

Through substitution, the integral becomes much more manageable, showcasing the method's power in solving complex integrals easily.

Integration by Parts

Integration by parts is another vital technique in calculus integration. It stems from the product rule for differentiation and helps when dealing with products of functions. Although not used in the original problem, understanding this method is crucial.

The rule is defined as:\[\int u \, dv = uv - \int v \, du\]Here's how to apply it:

• Choose \( u \) and \( dv \) from the integrand. Ideally, \( u \) is a function that becomes simpler upon differentiation, and \( dv \) is easy to integrate.
• Differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \).
• Substitute in the formula to simplify the initial integral.

While the original problem utilized substitution, integrating \( \int x \ln(x) \, dx \) would be perfect for integration by parts, illustrating its relevance and importance to different types of integrals.

Definite and Indefinite Integrals

Understanding definite and indefinite integrals is essential for calculus students. They both involve finding antiderivatives but differ in their application.Indefinite integrals refer to the family of functions representing the antiderivative of a given function. They include a constant \( C \) because integration involves undoing differentiation, which could have lost this constant along the way.

Definite integrals, denoted with limits, calculate the net area under a curve between two points. A definite integral from \( a \) to \( b \), \( \int_{a}^{b} f(x) \, dx \), results in a specific number, the area between the \( x \)-axis and \( f(x) \) from \( a \) to \( b \). Here's the main distinction:

• Indefinite Integral: Expressed as \( \int f(x) \, dx = F(x) + C \)
• Definite Integral: Calculated with limits as \( F(b) - F(a) \).

In the context of our solved problem, we dealt with an indefinite integral resulting in a generalized solution \( -\cos(x^2) + C \). It's essential to note both types, as they serve different mathematical and real-world applications.