Problem 28 - Calculus Single Variable

Question

Use the method of substitution to evaluate the definite integrals. $$ \int_{0}^{3} \frac{\sin (\pi \sqrt{t+1})}{\sqrt{t+1}} d t $$

Step-by-Step Solution

Answer

The evaluated definite integral is \(-\frac{4}{\pi}\).

Step 1: Choose a Substitution

In the given problem, we have the expression \(\sin (\pi \sqrt{t+1})\) and its derivative, \(\frac{1}{\sqrt{t+1}}\). Let's use the substitution \(u = \sqrt{t+1}\), which implies \(t = u^2 - 1\).

Step 2: Differentiate the Substitution

Differentiate \(u = \sqrt{t+1}\) with respect to \(t\): \(\frac{du}{dt} = \frac{1}{2\sqrt{t+1}} = \frac{1}{2u}\). Therefore, \(dt = 2u \, du\).

Step 3: Change the Integration Limits

Substitute the limits of \(t\) into \(u = \sqrt{t+1}\). When \(t = 0\), \(u = \sqrt{1} = 1\). When \(t = 3\), \(u = \sqrt{4} = 2\). Hence, the new limits are from \(u = 1\) to \(u = 2\).

Step 4: Substitute and Transform the Integral

Substituting in the integral, we get:\[\int_{1}^{2} \sin(\pi u) \cdot 2u \cdot \frac{1}{u} \, du = 2 \int_{1}^{2} \sin(\pi u) \, du\]This simplifies to \(2 \int_{1}^{2} \sin(\pi u) \, du\).

Step 5: Integrate the Transformed Integral

Integrate \(\int \sin(\pi u) \, du\):\[\int \sin(\pi u) \, du = -\frac{1}{\pi} \cos(\pi u) + C\]Thus, we have:\[2 \left[-\frac{1}{\pi} \cos(\pi u) \right]_{1}^{2} = -\frac{2}{\pi} \left[\cos(2\pi) - \cos(\pi)\right]\]

Step 6: Evaluate the Definite Integral

Evaluate the cosine terms:\(\cos(2\pi) = 1\text{ and }\cos(\pi) = -1\)So, the expression becomes:\[-\frac{2}{\pi} (1 - (-1)) = -\frac{2}{\pi} \times 2 = -\frac{4}{\pi}\]

Key Concepts

Method of Substitution

The Method of Substitution, often called the "u-substitution," is a powerful technique used to simplify the integral of a complex function. Here's how it works:

• First, identify a part of the integrand, the function inside the integral, which if replaced, simplifies the process. For example, in our problem, the expression \( \sin(\pi \sqrt{t+1}) \) can seem daunting at first.
• Next, assign it to a new variable. In this case, we substitute \( u = \sqrt{t+1} \), making differentiation and subsequent calculations easier.
• The goal of this technique is to transform the integral into a basic form that is more easily solved. Once the math is done, remember to also adjust for the limits of integration if dealing with definite integrals.

By applying this method, the original integral's complexity is simplified, enabling us to thoroughly solve the problem, step-by-step.

Integration by Substitution

Integration by substitution is a specific technique that stems from the basic principle of substitution. It is especially useful when direct integration is difficult or impossible.Here is the process:

• The first step involves selecting a substitution that will simplify the integral. Here, \( u = \sqrt{t+1} \) simplifies both \( \sin(\pi \sqrt{t+1}) \) and its derivative.
• The second step requires us to find the derivative of the chosen substitution: \( \frac{du}{dt} = \frac{1}{2u} \).
• Replace \( dt \) with the expression found from this derivative: here, \( dt = 2u \, du \).
• Resubstitute these back into the integral which now becomes simple enough to integrate directly.

By redefining the integral with these steps, the problem morphs into one that aligns with standard integral tables or basic integration rules.

Trigonometric Integration

When we incorporate trigonometric functions like \( \sin \) into integrals, it introduces the concept of trigonometric integration. It requires special attention because trigonometric functions have unique properties.In this exercise:

• After substitution, the integral simplifies to \( 2\int \sin(\pi u) \, du \).
• The function \( \sin \) often simplifies after integrating, here, using the identity \( \int \sin(x) \, dx = -\cos(x) + C \).
• Notice how the integration results in function transformations, requiring us to keep constants, such as \( \frac{1}{\pi} \), in mind as they modify the integral's result.

Understanding these particular characteristics is crucial, especially when solving problems incorporating sine, cosine, or other trigonometric functions.

Definite Limits

Definite integrals are defined between two specific limits that denote an interval over the domain of integration, giving a numeric result.In our problem, transitioning from indefinite to definite integrals involves:

• Determining new limits of integration. For the new variable \( u = \sqrt{t+1} \), the limits adjust from \( t = 0, 3 \) to \( u = 1, 2 \).
• After integrating the function \( \sin(\pi u) \), use these limits from \( u = 1 \) to \( u = 2 \) to evaluate the integral. This results in computing the difference between the integrated function evaluated at these bounds.
• Here, we calculate: \( 2 [-\frac{1}{\pi} \cos(\pi u)]_{1}^{2} \), resulting in a specific numerical value.

Definite limits offer precision, converting an algebraic expression into a concrete numerical conclusion.