Problem 29 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int \sin (\ln x) d x $$

Step-by-Step Solution

Answer

\( \int \sin(\ln x) \, dx = \frac{1}{2}(x \sin(\ln x) - x \cos(\ln x)) + C \).

Step 1: Identify parts for integration by parts formula

The integration by parts formula is: \( \int u \, dv = uv - \int v \, du \). For the integral \( \int \sin(\ln x) \, dx \), assign \( u = \sin(\ln x) \) and \( dv = dx \). This means we will need to differentiate \( u \) and integrate \( dv \).

Step 2: Differentiate and Integrate

Differentiate \( u = \sin(\ln x) \) to find \( du \). Using the chain rule, \( du = \cos(\ln x) \cdot \frac{1}{x} \, dx \). Integrate \( dv = dx \) to get \( v = x \).

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \cdot \cos(\ln x) \cdot \frac{1}{x} \, dx. \] Simplify the remaining integral: \[ x \cdot \cos(\ln x) \cdot \frac{1}{x} = \cos(\ln x), \] so the equation becomes: \[ \int \sin(\ln x) \, dx = x \sin(\ln x) - \int \cos(\ln x) \, dx. \]

Step 4: Solve the Simplified Integral

Notice that \( \int \cos(\ln x) \, dx \) has the same form as the original problem. Reapply integration by parts on this integral with \( u = \cos(\ln x) \) and \( dv = dx \). Find \( du = -\sin(\ln x) \cdot \frac{1}{x} \, dx \) and \( v = x \).

Step 5: Apply Integration by Parts Again

Using integration by parts again, \[ \int \cos(\ln x) \, dx = x \cos(\ln x) + \int \sin(\ln x) \, dx. \] The equation now reads \[ \int \sin(\ln x) \, dx = x \sin(\ln x) - (x \cos(\ln x) + \int \sin(\ln x) \, dx). \] Simplify and rearrange to find \[ 2 \int \sin(\ln x) \, dx = x \sin(\ln x) - x \cos(\ln x), \] so finally, \[ \int \sin(\ln x) \, dx = \frac{1}{2}(x \sin(\ln x) - x \cos(\ln x)). \]

Step 6: Conclude with the Result

So the integral of \( \sin(\ln x) \) is: \[ \int \sin(\ln x) \, dx = \frac{1}{2} (x \sin(\ln x) - x \cos(\ln x)) + C, \] where \( C \) is the constant of integration.

Key Concepts

Chain Rule

The chain rule is a fundamental method in calculus for finding the derivative of composite functions. If you have a function inside another function, typically represented as \( f(g(x)) \), the chain rule helps differentiate it.
For instance, to differentiate \( \sin(\ln x) \), we identify the outer function as \( f(u) = \sin u \) where \( u = \ln x \).
The chain rule formula is:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

This means we differentiate the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function. In this case:

• \( \frac{d}{dx}[\sin(\ln x)] = \cos(\ln x) \cdot \frac{1}{x} \)

This rule allows us to handle more complex differentiation problems effectively.

Differentiation

Differentiation is the process of finding the derivative, which represents the rate of change of a function as its input varies.
The derivative provides insight into the function's behavior and can help find slopes of tangent lines, optimization problems, and more.
When differentiating \( \sin(\ln x) \) during integration by parts, using the chain rule as a tool helps understand its derivative.
In general, knowing the basic differentiation formulas like those for power functions, exponential functions, and trigonometric functions can be immensely helpful.

• For example, \( \frac{d}{dx}[x^n] = nx^{n-1} \).
• Another example is \( \frac{d}{dx}[e^x] = e^x \).
• For trigonometric functions, like sine and cosine, \( \sin(x) \) differentiates to \( \cos(x) \), and \( \cos(x) \) differentiates to \(-\sin(x) \).

Such rules are essential tools in the calculus toolkit for effectively tackling problems.

Integration

Integration is the reversal of differentiation. While differentiation finds the rate of change, integration focuses on finding the total accumulation or area under a curve.
In this exercise, we use integration by parts, a specific technique that helps integrate products of functions where one function is easily differentiated, and the other is easily integrated.
The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

Choose \( u \) to be a function that's easily differentiated, and \( dv \) as something that's easily integrated.
In our example, with \( \int \sin(\ln x) \, dx \), choosing \( u = \sin(\ln x) \) and \( dv = dx \) eventually leads to simplification. This breaks down into reverse differentiation followed by straightforward integration.

Definite Integral

A definite integral calculates the accumulation of quantities and the net area between the curve and the x-axis over a specific interval. Unlike indefinite integrals, which include a constant of integration, definite integrals result in specific numerical values.
For definite integrals, the notation \( \int_a^b f(x) \, dx \) signifies integrating the function \( f(x) \) from \( a \) to \( b \), where \( a \) and \( b \) are the limits of integration.
The Fundamental Theorem of Calculus reveals a connection between differentiation and integration by stating that if \( F \) is an antiderivative of \( f \), then:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

This theorem allows practitioners to compute definite integrals by evaluating the antiderivative at the endpoints and subtracting the results. Emphasizing its usefulness, definite integrals model real-life problems, including finding area, displacement, and total accumulated change in various contexts.