Problem 30 - Calculus

Question

Use the method of substitution to find each of the following indefinite integrals. $$ \int x^{6}\left(7 x^{7}+\pi\right)^{8} \sin \left[\left(7 x^{7}+\pi\right)^{9}\right] d x $$

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{63} \cos((7x^7 + \pi)^9) + C\).

Step 1: Choose a substitution

Let \( u = (7x^{7} + \pi)^9 \). This will help simplify the integral. We choose this substitution because the integral involves a complicated function raised to a power and another function (sine) applied to the same expression.

Step 2: Differentiate the substitution

Differentiate \( u = (7x^{7} + \pi)^9 \) with respect to \( x \) to find \( du \). The derivative is given by:\[ du = 9(7x^7 + \pi)^8 imes 7x^6 \, dx = 63x^6(7x^7 + \pi)^8 \, dx \]

Step 3: Solve for \( dx \)

From the equation \( du = 63x^6(7x^7 + \pi)^8 \, dx \), solve for \( dx \): \[ dx = \frac{1}{63x^6(7x^7 + \pi)^8} \, du \]

Step 4: Substitute into the integral

Substitute \( u \) and \( dx \) into the integral:\[ \int x^6 (7x^7 + \pi)^8 \sin((7x^7 + \pi)^9) \, dx = \int \sin(u) \, \frac{1}{63} \, du \]

Step 5: Solve the new integral

The new integral \( \int \sin(u) \, \frac{1}{63} \, du \) simplifies to:\[ \frac{1}{63} \int \sin(u) \, du = -\frac{1}{63} \cos(u) + C \] where \( C \) is the constant of integration.

Step 6: Back-substitute

Replace \( u \) with \( (7x^7 + \pi)^9 \) in the antiderivative obtained:\[ -\frac{1}{63} \cos(u) + C = -\frac{1}{63} \cos((7x^7 + \pi)^9) + C \]

Key Concepts

Indefinite Integrals

Indefinite integrals are the opposite of derivatives in calculus. They are used to find a function from its derivative, essentially allowing us to work backward from a rate of change to the original function. An indefinite integral is represented by the integral symbol \( \int \) and includes a constant of integration, typically denoted as \( C \). This constant accounts for any number that could have been in the original function before differentiation.

If you encounter the function \( F(x) \), whose derivative is \( f(x) \), the indefinite integral of \( f(x) \) is often written as \( F(x) + C \), implying that \( \int f(x) \, dx = F(x) + C \). When working with indefinite integrals, you're identifying a set of functions with the same derivative, allowing flexible solutions in integration problems. They play an essential role in solving practical problems in physics, engineering, and economics by finding quantities given their rates of change.

For example, the step-by-step solution of the given integral demonstrates knowing how to integrate a function with a complicated expression by employing the substitution method. Indefinite integrals can appear complex, but with practice and knowledge of techniques, solving them becomes manageable.

Integration Techniques

Integration techniques are methods that help solve integrals. Various techniques can make solving integrals easier, especially when dealing with complex or unfamiliar expressions. The substitution method, used in the given problem, is a powerful technique. It simplifies integrals by transforming difficult expressions into a more manageable form.

Some popular integration techniques include:

• Substitution Method: Ideal when the integrand involves a composite function. It involves changing the variable to simplify.
• Integration by Parts: Useful when the integrand is a product of functions. It leverages the product rule from differentiation to facilitate the integral.
• Partial Fractions: Applicable when the integrand is a rational function. This technique decomposes a complex fraction into simpler parts.
• Trigonometric Substitutions: This uses trigonometric identities to simplify integrals that involve radicals.

In our exercise, the substitution method dramatically simplified the problem by setting \( u = (7x^7 + \pi)^9 \). After substituting, the original complex integral transformed into a simpler trigonometric integral of sin(u).

Mastery of these techniques increases efficiency in solving integrals and extends one's ability to tackle a wide range of calculus problems.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are fundamental building blocks in mathematics, particularly in calculus and integration. These functions appear frequently in various real-life applications, such as physics, engineering, and signal processing.

In the context of integration, recognizing trigonometric functions within integrals is crucial, as it guides the choice of techniques for solving them. In the given exercise, the function \( \sin((7x^7 + \pi)^9) \) was presented inside the integral. This highlights a classic case where the sine function's behavior can be utilized after substitutions to achieve integration more smoothly.
Some key properties of trigonometric functions include:

• They have periodic behavior and repeat their values at regular intervals.
• They can be represented using the unit circle, which provides a geometrical perspective for their values.
• They form the basis for defining other trigonometric identities and functions, which can assist in simplifying and solving integrals.

Recognizing when trigonometric functions can be simplified using identities, combined with substitution techniques, opens up a range of straightforward solutions to what initially might seem like very complicated integration problems. Understanding these functions deeply enables more effective problem-solving skills in calculus.