Problem 19

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int \sin (6 x-7) d x $$

Step-by-Step Solution

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Answer

The integral is $-\frac{1}{6} \cos(6x - 7) + C$.

1Step 1: Choose a Substitution

Let's choose a substitution that simplifies the integrand. We set $ u = 6x - 7 $.

2Step 2: Differentiate the Substitution

Differentiating $ u = 6x - 7 $ with respect to $ x $, we get $ \frac{du}{dx} = 6 $. Therefore, we can express $ dx $ as $ dx = \frac{du}{6} $.

3Step 3: Rewrite the Integral in Terms of 'u'

Substitute $ u $ and $ dx $ in the original integral: \[ \int \sin(u) \cdot \frac{1}{6} \, du \]. This simplifies to $ \frac{1}{6} \int \sin(u) \, du $.

4Step 4: Integrate with Respect to 'u'

Find the integral of $ \sin(u) $ which is $ -\cos(u) $. Therefore, \[ \frac{1}{6} \int \sin(u) \, du = -\frac{1}{6} \cos(u) + C \], where $ C $ is the constant of integration.

5Step 5: Substitute Back the Original Variable

Replace $ u $ with the original expression $ 6x - 7 $. Thus, \[ -\frac{1}{6} \cos(6x - 7) + C \].

Key Concepts

Substitution MethodTrigonometric IntegralsIntegration Techniques

Substitution Method

The substitution method is a popular technique used in calculus to solve certain types of integrals, especially indefinite integrals. It's a way to simplify the integration process by transforming the integrand into a simpler form. This method is comparable to the reverse action of the chain rule in differentiation.

Here's how the substitution method works:

First, identify a part of the integrand that can be substituted by a single variable, often denoted as "u".
Choose this substitution such that it simplifies the expression, making it easier to integrate.
Differentiating this substitute expression with respect to the original variable gives the derivative, often rearranged to substitute the differential (e.g., "dx") in the integrand.
Rewrite the entire integrand in terms of the new variable "u", making it easier to evaluate the integral.
After integration, substitute back the original variable expression.

In our example, the substitution chosen was "u = 6x - 7," turning the integral into one involving a simple trigonometric function, "sin(u)".

Trigonometric Integrals

Trigonometric integrals are those that involve sine, cosine, and other trigonometric functions. These integrals often require special techniques or substitutions to evaluate.

When integrating trigonometric functions, recall the basic derivatives and antiderivatives of sine and cosine:

The derivative of $ \sin(x) $ is $ \cos(x) $, so the integral of $ \sin(x) $ is $ -\cos(x) + C $.
The derivative of $ \cos(x) $ is $ -\sin(x) $, so the integral of $ \cos(x) $ is $ \sin(x) + C $.

Knowing these connections helps to find integrals of functions like $ \sin(u) $. In our example, the substitution method allowed for simplifying the integral to $ \int \sin(u) \, du $, which directly integrates to $ -\cos(u) + C $. This step emphasizes the importance of recognizing and using these basic trigonometric identities in integration.

Integration Techniques

Integration techniques are various methods used to find integrals, accommodating different kinds of integrands. These methods are crucial for dealing with complex expressions that cannot simply be integrated using basic calculus rules.

Some common integration techniques include:

Substitution Method: As already explained, it simplifies complex expressions by changing variables.
Integration by Parts: This technique is typically used for products of functions and is based on the product rule for differentiation.
Partial Fraction Decomposition: Useful for rational functions, where the integrand is a ratio of polynomials.
Trigonometric Substitution: Helpful for integrands involving square roots and trigonometric identities.

In this exercise, the substitution method was the most appropriate choice, highlighting how different techniques suit different types of functions. Mastery of various techniques enriches problem-solving skills, ensuring preparedness for a wide range of integrals.

Problem 19

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