Problem 32

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int x^{6} \sin \left(3 x^{7}+9\right) \sqrt[3]{\cos \left(3 x^{7}+9\right)} d x $$

Step-by-Step Solution

Verified

Answer

$-\frac{3}{84} \cdot \cos^{4/3}(3x^7 + 9) + C$

1Step 1: Identify substitution

Let's identify a substitution that simplifies the integral. In this case, let $ u = 3x^7 + 9 $. This substitution is chosen because both the $ \sin(u) $ and the $ \cos(u) $ terms inside the integral depend on this expression.

2Step 2: Differentiate the substitution

Differentiate $ u $ with respect to $ x $ to find $ dx $.\[du = 21x^6 \, dx\]Rearrange to express $ dx $:\[dx = \frac{1}{21x^6} \, du\]

3Step 3: Substitute into the integral

Substitute $ u $ and $ dx $ back into the integral, replacing $ 3x^7 + 9 $ with $ u $ and $ dx $ with $ \frac{1}{21x^6} \, du $. The integral becomes:\[\int x^6 \sin(u) \sqrt[3]{\cos(u)} \cdot \frac{1}{21x^6} \, du\]

4Step 4: Simplify the integral

Notice that $ x^6 $ terms cancel each other out, simplifying the integral to:\[\frac{1}{21} \int \sin(u) \sqrt[3]{\cos(u)} \, du\]

5Step 5: Solve the simplified integral

Now, integrate $ \sin(u) \sqrt[3]{\cos(u)} $ with respect to $ u $. The integral is \[ -\frac{3}{4} \cdot \cos^{4/3}(u) + C \].

6Step 6: Substitute back for $ u $

Substitute $ u = 3x^7 + 9 $ back into the integrated result. The solution to the integral becomes:\[-\frac{3}{4} \cdot \cos^{4/3}(3x^7 + 9) + C\]

Key Concepts

Method of SubstitutionTrigonometric IntegralsIntegral CalculusU-Substitution

Method of Substitution

The method of substitution is a powerful technique in integral calculus used to simplify integrals by changing variables. It essentially "transforms" a complex integral into a more manageable one.
The key idea is to identify a part of the integral as a new variable, substitute, and then express everything in terms of this new variable. This not only simplifies the process but also helps in recognizing standard integral forms.
When applying substitution, it's important to:

Identify a suitable substitution that simplifies the integral, usually the innermost function or expression.
Differentiate the substitution to find the derivative and rewrite the original differential as a function of the new variable.
Fully substitute both the function and the differential to convert the entire integral into a new variable.
Reintegrate and, finally, substitute back the original expressions to get the answer in terms of the original variable.

Trigonometric Integrals

Trigonometric integrals involve trigonometric functions like sine, cosine, and others. They often appear complicated due to these functions' periodic and oscillating nature, but they can be simplified significantly with substitution.
When performing substitution for trigonometric integrals, it helps to choose a substitution that turns the trigonometric expressions into more standard forms. For example, expressing angles or complicated expressions in the argument of sine and cosine functions as a single variable.
For integrals involving expressions like $ \sin(u) $ and $ \cos(u) $, pay attention to:

How trigonometric identities can play a role in simplifying parts of the integral after substitution.
Whether substitution helps to reduce the degree of these trigonometric functions to standard forms.
Recognize common integral patterns that result from these trigonometric functions.

Integral Calculus

Integral calculus focuses on the concepts of integration and finding antiderivatives. It's the counterpart to differential calculus, which is concerned with the derivative.
In integral calculus, indefinite integrals represent a family of functions whose derivative is the function being integrated. When solving an indefinite integral, you aim to find a formula that describes all potential antiderivatives.
Key features in integral calculus important for solving integrals include:

Understanding the properties of definite and indefinite integrals.
Using substitution to turn complex integrals into simpler ones.
Recognizing and applying integration rules, such as the power rule, chain rule, and integration by parts.

Typically, the solution of an indefinite integral includes a constant of integration, $ C $, representing that there are infinitely many specific antiderivatives.

U-Substitution

U-substitution is a specific application of the substitution method where you replace the given function with a new variable, $ u $. This is especially helpful when dealing with integrals that involve composite functions.
In u-substitution:

Select part of the integrand, usually the inner function of a composite, and set it equal to $ u $.
Find the differential $ du $ and rewrite $ dx $ in terms of $ du $. This makes changing variables possible within the integral.
Completely replace the original variables with the new ones to simplify the integral.
Integrate using this new variable, making it typically easier to perform standard integration.
Finally, reverse the substitution to express your final answer in terms of the original variables.

U-substitution often resembles reversing the chain rule from differentiation, making it an essential tool in solving complicated integrals.

Problem 31

Problem 32

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