Problem 40 - Calculus for Biology and Medicine

Question

First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals. $$ \int \sin \sqrt{x} d x $$

Step-by-Step Solution

Answer

\(-2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C\)

Step 1: Make a Substitution

Start by letting \( u = \sqrt{x} \). This implies \( x = u^2 \). Differentiate both sides with respect to \( x \) to find \( dx \) in terms of \( du \): \( \frac{d}{dx} (u) = \frac{1}{2\sqrt{x}} = \frac{1}{2u} \). Thus, \( dx = 2u \, du \). Substitute \( u \) for \( \sqrt{x} \) and \( 2u \, du \) for \( dx \) in the integral.

Step 2: Rewrite the Integral

The integral now becomes \( \int \sin(u) \cdot 2u \, du \), or \( 2 \int u \cdot \sin(u) \, du \). This sets the stage for integration by parts.

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Select \( u = u \) and \( dv = \sin(u) \, du \), which means \( du = du \) and \( v = -\cos(u) \) after integrating \( dv \).

Step 4: Execute Integration by Parts

Apply the integration by parts: \( 2 \left( -u \cos(u) - \int -\cos(u) \, du \right) \). Simplify this to: \( 2 \left( -u \cos(u) + \sin(u) \right) \).

Step 5: Substitute Back

Replace \( u \) with \( \sqrt{x} \) to convert back in terms of \( x \). The solution is: \( -2\sqrt{x}\cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C \), where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is used to simplify a complex integral by turning it into a more manageable form. It's particularly useful when dealing with integrals involving composite functions, such as \( \sin \sqrt{x} \).To use substitution effectively, make a clever choice of substitution. In this problem, we chose \( u = \sqrt{x} \), which directly impacts the complexity of the integral. As a result, \( x \) becomes \( u^2 \), which allows us to convert the variables and the differential. By differentiating \( u = \sqrt{x} \), we find \( dx = 2u \, du \).

• Change the variable: \( u = \sqrt{x} \)
• Differentiate to find \( dx \) in terms of \( du \)
• Substitute \( u \) and \( du \) in the integral

This transforms the integral from its more complex original form to a simpler one: \( \int \sin(u) \cdot 2u \, du \), thus setting the scene for applying another method like integration by parts.

Indefinite Integrals

Indefinite integrals (also known as antiderivatives) result in a general form of the original function before differentiation, plus a constant \( C \). Unlike definite integrals, which yield a numerical value, indefinite integrals represent a family of functions that differ by a constant. The goal is to find the antiderivative of the function we are integrating.
In our problem, we applied the substitution method first to simplify the given integral. This helped in setting up integration by parts to evaluate the simpler integral. Following the integration process:

• Recognize the integral form: \( \int \sin(u) \cdot 2u \, du \).
• Apply integration by parts: select parts \( u = u \) and \( dv = \sin(u) \, du \).
• Integrate and simplify: Use the formula \( \int u \, dv = uv - \int v \, du \).

After finding the integral, we replaced our substitute variable \( u \) back to \( \sqrt{x} \), adding \( C \) to signify all possible antiderivatives.

Trigonometric Integrals

Trigonometric integrals involve functions like \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and their combinations. They are common in calculus and often require strategies such as substitution or integration by parts.
In this specific problem, the integral involved \( \sin \sqrt{x} \). After using substitution, we expressed our integral in terms of \( \sin(u) \). Here's how trigonometric integrals come into play:

• The trigonometric part, \( \sin(u) \), is smoother to handle after substitution.
• With \( dv = \sin(u) \, du \), the integration part gives us \( v = -\cos(u) \).
• This part is then integrated in a straightforward manner.

Trigonometric identities and properties of integrals make solving these integrals more direct and manageable. In our solution, converting \( \sin \sqrt{x} \) to a simpler problem by substitution was key to using the familiar form \( \int \sin(u) \, du \).