Problem 14 - Applied Calculus

Question

Find the integrals. $$\int(\theta+1) \sin (\theta+1) d \theta$$

Step-by-Step Solution

Answer

The integral is \(- (\theta + 1) \cos(\theta + 1) + \sin(\theta + 1) + C\)."

Step 1: Apply Substitution

To simplify the integration process, let's use substitution. Set \( u = \theta + 1 \). Then, the derivative \( du = d\theta \). This substitution will change the integral to \( \int u \sin u \, du \).

Step 2: Integration by Parts

Apply integration by parts, where \( \int u \, dv = uv - \int v \, du \). Here, choose \( v = -\cos u \) and \( du = du \), while \( u = u \) and \( dv = \sin u \, du \). Apply the formula: \( uv - \int v \, du = -u \cos u + \int \cos u \, du \).

Step 3: Integrate Remaining Function

Now, integrate \( \int \cos u \, du \), which is \( \sin u \). Substitute back into the equation: \( -u \cos u + \int \cos u \, du = -u \cos u + \sin u \).

Step 4: Substitute Back to Original Variable

Revert to the original variable \( \theta \) by substituting back \( u = \theta + 1 \). The expression becomes \( - (\theta + 1) \cos(\theta + 1) + \sin(\theta + 1) + C \), where \( C \) is the constant of integration.

Key Concepts

Integration by Substitution

In integral calculus, integration by substitution is a method used to simplify complex integrals. This technique involves changing the variable to transform the integral into an easier form. You start by choosing a substitution that will simplify the integral. For example, in the integral \( \int(\theta + 1) \sin (\theta + 1) \, d\theta \), the substitution \( u = \theta + 1 \) is chosen.
This means the derivative \( du = d\theta \), and the integral becomes \( \int u \sin u \, du \). The substitution reduces the complexity and allows for an easier evaluation using other methods.
When choosing a substitution, aim for an expression in the integral that, when substituted, results in simpler factors or cancels out. This process requires practice to recognize helpful substitutions.

• Identify the part of the integral to replace.
• Substitute with a new variable.
• Determine the new differential \( du \).
• Replace and simplify the integral.

In this exercise, substitution is the first crucial step to solve the integral effectively.

Integration by Parts

Integration by parts is another essential technique in integral calculus, especially useful when integrating the product of two functions. The formula is derived from the product rule for differentiation and is expressed as \( \int u \, dv = uv - \int v \, du \).
To apply integration by parts, carefully choose your \( u \) and \( dv \). For the integral \( \int u \sin u \, du \), let \( u = u \) and \( dv = \sin u \, du \).
This gives \( v = -\cos u \) after integrating \( dv \), and the equation for integration by parts becomes \(-u \cos u + \int \cos u \, du \). The integrity of this method lies in transforming the given integral into a simpler form that can be directly solved.

• Select \( u \) such that its derivative \( du \) simplifies the expression.
• The choice of \( dv \) should be something easy to integrate.
• Apply the integration by parts formula.
• Simplify the expressions, if possible.

In our solution, integration by parts helps streamline the calculation toward a solvable endpoint.

Trigonometric Integration

Trigonometric integration involves integrals that include trigonometric functions such as sine, cosine, or tangent. Understanding these functions and their integrals is key to solving a variety of calculus problems. In this exercise, after simplifying the integral with substitution and integrating by parts, we encounter \( \int \cos u \, du \).
The integral \( \int \cos u \, du \) simplifies to \( \sin u \). Recognizing basic trigonometric integrals can save time and simplify the process. Familiar functions and their integrals are:

• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)
• \( \int \tan x \, dx = -\ln |\cos x| + C \)

Knowing these foundational integral forms is very beneficial.
Even when not initially obvious, looking for ways to apply trigonometric identities or substitutions can often make a seemingly difficult integral manageable. Once you find the antiderivative, always remember to return to the original variable to complete the process.