Question

For the following exercises, use this information: The inner product of two functions \(f\) and \(g\) over \([a, b]\) is defined by \(f(x) \cdot g(x)=\langle f, g\rangle=\int_{a}^{b} f \cdot g d x\). Two distinct functions \(f\) and \(g\) are said to be orthogonal if \(\langle f, g\rangle=0\). Integrate \(y^{\prime}=\sqrt{\tan x} \sec ^{4} x\)

Short answer

\( y(x) = \frac{2}{3} (\tan x)^{3/2} + \frac{4}{5} (\tan x)^{5/2} + \frac{2}{7} (\tan x)^{7/2} + C \)

Step-by-step solution

Understand the Integration Problem

The task is to integrate the given derivative function: \[ y' = \sqrt{\tan x} \sec^4 x \]This means finding the antiderivative or the original function \( y(x) \).

Substitute to Simplify the Integral

We need to make a substitution to simplify the integral. Let \( u = \tan x \). Therefore, the derivative \( du = \sec^2 x \, dx \) which can be used for substitution.

Express Everything in Terms of u

Rewrite the integral with the substitution made:\[ y = \int \sqrt{u} \sec^4 x \, dx \]Since \( \sec^2 x = 1 + \tan^2 x \), we have\[ \sec^4 x = (1 + u)^2 \sec^2 x \]Therefore, the integral becomes:\[ y = \int \sqrt{u} (1+u)^2 \ du \] We use \( du = \sec^2 x \, dx \) from the substitution.

Expand and Integrate

Expand \( (1 + u)^2 \) to simplify:\[ (1 + u)^2 = 1 + 2u + u^2 \]Thus, the integral becomes:\[ y = \int (\sqrt{u} + 2u^{3/2} + u^{5/2}) \, du \]Now, integrate each term separately.

Solve the Integral

The integration of each term gives:1. \( \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C_1 \)2. \( \int 2u^{3/2} \, du = 2 \cdot \frac{2}{5} u^{5/2} + C_2 = \frac{4}{5} u^{5/2} + C_2 \)3. \( \int u^{5/2} \, du = \frac{2}{7} u^{7/2} + C_3 \)Therefore,\[ y = \frac{2}{3} u^{3/2} + \frac{4}{5} u^{5/2} + \frac{2}{7} u^{7/2} + C \]

Substitute Back to x

Now that we have the antiderivative in terms of \( u \), substitute back \( u = \tan x \):\[ y(x) = \frac{2}{3} (\tan x)^{3/2} + \frac{4}{5} (\tan x)^{5/2} + \frac{2}{7} (\tan x)^{7/2} + C \]

Conclusion

We have integrated the function and expressed \( y(x) \) in terms of \( x \):\[ y(x) = \frac{2}{3} (\tan x)^{3/2} + \frac{4}{5} (\tan x)^{5/2} + \frac{2}{7} (\tan x)^{7/2} + C \]

Key concepts

Orthogonal Functions

In mathematics, functions are said to be orthogonal when they are perpendicular to each other in the context of function space. This idea is similar to how two vectors can be perpendicular in an elementary geometry context. Orthogonality in functions is determined by calculating the inner product, and if it returns zero, the functions are orthogonal.

For two functions, say \( f(x) \) and \( g(x) \), defined on the interval \([a, b]\), they are orthogonal if their inner product, given by \( \langle f, g \rangle = \int_a^b f(x) g(x) \, dx \), equals zero. This concept is particularly useful in fields like signal processing, where orthogonal functions represent unique, non-overlapping components of signals.

Orthogonal functions play a fundamental role in solving differential equations, optimizing functions, and decomposing signals. Understanding them helps in harnessing the full potential of various mathematical methods in engineering and physics.

Inner Product of Functions

The inner product of functions is a way to multiply two functions over an interval and results in a scalar (a single number). This operation measures the degree of overlapping between the two functions. A crucial part of functional analysis, the inner product extends the notion of the dot product of vectors to the function space.

Given two functions \( f(x) \) and \( g(x) \), their inner product on the interval \([a, b]\) is given by:

• \( \langle f, g \rangle = \int_a^b f(x) g(x) \, dx \)

This calculation is commonly involved in determining orthogonality and projecting functions.

In practice, the idea of the inner product reveals insights about the functions’ behavior, comparability, and can aid in the computation of coefficients in series expansions like Fourier Series. It is a stepping stone in more advanced mathematical methods, making it a significant concept to grasp.

Substitution Method in Integration

The substitution method is a powerful technique in integration that eases solving complex integrals. It involves changing variables to simplify the integral into an easier form. Often, this method can directly help in evaluating integrals that seem cumbersome or unsolvable at first glance.

To perform substitution, follow these basic steps:

• Identify a part of the integrand that can be substituted, typically by setting \( u = g(x) \) for some function \( g \).
• Find \( du \) from the derivative of \( u = g(x) \).
• Express the integral in terms of \( u \), making it easier to solve.
• Integrate the simpler expression.
• Substitute back the original variable to express the integral’s result in terms of the original variable.

The substitution technique is akin to the reverse of the chain rule from differentiation. It proves invaluable when facing polynomials, trigonometric integrals, and logarithmic functions, making it indispensable in calculus education.

Antiderivative

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. When you have a derivative \( y' \), the task is to find a function \( y(x) \) whose derivative is \( y' \).

This process is called finding the antiderivative, and it's depicted as:

• \( y(x) = \int f(x) \, dx = F(x) + C \)

Here, \( F(x) \) is one antiderivative of \( f(x) \) and \( C \) denotes the constant of integration. This constant is essential because when you differentiate a constant, it disappears, meaning multiple functions can have the same derivative, differing only by a constant.

Knowing how to find antiderivatives opens up solving solutions to differential equations and understanding the accumulation of quantities over time or space. Mastery of finding antiderivatives is essential for progressing in calculus and related fields.