Question

Integral of sec \(^{3} x\) Use integration by parts to show that $$\int \sec ^{3} x d x=\frac{1}{2} \sec x \tan x+\frac{1}{2} \int \sec x \, d x$$.

Short answer

Question: Show that the integral of \(\sec^3{x}\) can be expressed as a sum of a function and another integral term, using integration by parts. Answer: The integral of \(\sec^3{x}\) can be expressed as \(\int \sec^3 x dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x dx\).

Step-by-step solution

Understand Integration by Parts Formula

Integration by parts is derived from the product rule of differentiation, and the formula for integration by parts is given by: $$\int u \, dv = uv - \int v \, du$$ where \(u\) and \(v\) are functions of \(x\), \(du\) is the derivative of \(u\), and \(dv\) is the derivative of \(v\). We will apply this formula to the integral \(\int \sec^3 x \, dx\).

Choose 'u' and 'dv'

For this problem, we want to choose \(u\) and \(dv\) such that the product of \(uv\) simplifies, and the remaining integral becomes simpler and easier to solve. We take: - \(u = \sec x\) - \(dv = \sec^2 x \, dx\) Now, we differentiate \(u\) and integrate \(dv\) to find \(du\) and \(v\) respectively.

Differentiate 'u' and Integrate 'dv'

Differentiating \(u=\sec x\) and expressing in terms of \(dx\), we get: $$du=\sec x \tan x \, dx$$ Integrating \(dv=\sec^2 x \, dx\), we get: $$v=\tan x$$

Substitute 'u', 'v', 'du', and 'dv' in Integration by Parts Formula

Now, we substitute the values of \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula: $$\int \sec^3 x \, dx = \int u \, dv = uv - \int v \, du$$ Replacing \(u\), \(v\), \(du\), and \(dv\) with the values obtained in step 3, we get: $$\int \sec^3 x \, dx = \sec x \tan x - \int \tan x (\sec x \tan x) dx$$

Simplify the Remaining Integral

The remaining integral can be simplified as follows: $$\int \tan x (\sec x \tan x) dx = \int \sec x \tan^2 x dx$$ Since \(\tan^2 x = \sec^2 x - 1\), we can rewrite the integral as $$\int \sec x \tan^2 x dx = \int \sec x (\sec^2 x - 1) dx$$

Separate the Integral into Individual Terms

We can now separate the integral into individual terms for simpler integration: $$\int \sec x \tan^2 x dx = \int \sec x (\sec^2 x - 1) dx = \int \sec^3 x dx - \int \sec x dx$$

Solve for the Original Integral

We can now solve for the original integral \(\int \sec^3 x dx\) with the following equation: $$\int \sec^3 x dx = \sec x \tan x - (\int \sec^3 x dx - \int \sec x dx)$$ Rearranging terms, we get: $$2 \int \sec^3 x dx = \sec x \tan x + \int \sec x dx$$ And finally, dividing both sides by 2, we get the desired result: $$\int \sec^3 x dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x dx$$

Key concepts

Integral of Secant Cubed

The integral of secant cubed, denoted as \( \int \sec^3 x \, dx \), is a unique trigonometric integral that requires a strategic approach to solve. It involves using the technique of integration by parts. An understanding of trigonometric identities is also essential, particularly the identity \( \tan^2 x = \sec^2 x - 1 \), which comes into play during the simplification process.

By choosing the appropriate parts to integrate and differentiate, the integral, at first glance complicated, is broken down into simpler components. We reach a point where the original integral, \( \int \sec^3 x \, dx \), reappears on the right-hand side of the equation, allowing us to rearrange and solve for the integral. This algebraic maneuver results in an expression that includes the integral of \( \sec x \), which can be computed separately. When all the pieces come together, we obtain the solution involving both a product of \( \sec x \), and \( \tan x \) and an antiderivative of \( \sec x \).

Product Rule of Differentiation

The product rule of differentiation is a fundamental tool in calculus for finding the derivative of the product of two functions. This rule is formally expressed as, for two functions \( u(x) \) and \( v(x) \), \( (uv)' = u'v + uv' \).

The product rule underlies the concept of integration by parts, which essentially reverse-engineers this rule to find integrals. When we're looking to integrate a function that is the product of two other functions, breaking it down with integration by parts, just like we would with the product rule in differentiation, is crucial. This technique not only simplifies the problem but also allows us to handle more complicated integrals, such as \( \int \sec^3 x \, dx \).

Antiderivatives

Antiderivatives, also known as indefinite integrals, are an essential concept in calculus. They represent the inverse operation to differentiation. The antiderivative of a function \( f \) is a function \( F \) such that \( F' = f \). The most common notation for antiderivatives is the integral sign without bounds, \( \int f(x) \, dx \), which signifies that there are an infinite number of antiderivatives, differing only by a constant.

An antiderivative can be simpler to find for some functions, like polynomials, while others, particularly trigonometric functions, require the use of advanced techniques like integration by parts or trigonometric identities. Understanding how to find antiderivatives is a stepping stone to solving definite integrals—the area under a curve—and to applying calculus to real-world problems.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions, which can sometimes be quite challenging. These integrals often require the use of trigonometric identities to rewrite the integrand in a more manageable form. For example, identities like \( \sin^2x + \cos^2x = 1 \) or \( \tan^2x + 1 = \sec^2x \) are instrumental in transforming the original integral into simpler components that can be integrated individually.

In the case of \( \int \sec^3 x \, dx \), this integral is an advanced example of a trigonometric integral. It involves multiple steps, including the use of integration by parts and the manipulation of trigonometric identities. Mastery of these trigonometric integrals is not only academically rewarding but also builds a foundation for tackling more complex integrals in various fields of science and engineering.