Problem 21 - Essential Calculus

Question

Find the integral involving secant and tangent. $$ \int \sec ^{3} \pi x d x $$

Step-by-Step Solution

Answer

The integral \( \int \sec^3 \pi x dx \) equals \( \frac{\pi}{2} \sec \pi x \tan \pi x + \frac{1}{2} \ln| \sec \pi x + \tan \pi x | + C \).

Step 1: Set up the integral by parts

Following the formula for integration by parts: \[ \int u dv = uv - \int v du \], set \( u = \sec x \) and \( dv = \sec^2 x dx \). Then, differentiate and integrate to find \( du = \sec x \tan x dx \) and \( v = \tan x \]. The integral now becomes: \[ \int \sec^3 x dx = \sec x \tan x - \int \tan^2 x \sec x dx\].

Step 2: Use the Pythagorean identity

Remembering that \( \tan^2 x = \sec^2 x - 1 \), substitute it in the integral to get: \[ \int \sec^3 x dx = \sec x \tan x - \int (\sec^2 x \sec x - \sec x) dx\]. This can be further simplified to: \[ \sec x \tan x - \int \sec^3 x dx + \int \sec x dx \]

Step 3: Solve for the integral

Rearranging the equation to isolate the unsolved integral, get: \[ \int \sec^3 x dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x dx \]. The integral of \( \sec x \) is \( \ln | \sec x + \tan x | \) and therefore, the integral of \( \sec^3 x dx \) is: \[ \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln | \sec x + \tan x | + C \].

Step 4: Substitute \( \pi x \) back into the function

Lastly, as the original function is \( \sec^3 \pi x \), substitute \( \pi x \) back into the function and multiply with \( \pi \) to account for chain rule. The final answer is: \[ \frac{\pi}{2} \sec \pi x \tan \pi x + \frac{1}{2} \ln | \sec \pi x + \tan \pi x | + C \].

Key Concepts

Integration by Parts

Integration by parts is a crucial technique in integral calculus used to solve integrals where other methods fail. It's particularly useful when dealing with products of functions. The formula is straightforward:

• \( \int u \, dv = uv - \int v \, du \)

Here's how it works: you choose one part of your integrand as \( u \) and the other as \( dv \). Then, you differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \). Substituting back into the equation helps solve the integral. For example, in the exercise, we set \( u = \sec x \) and \( dv = \sec^2 x \, dx \). This led us through a series of steps to eventually find our integral solution.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables. They are vital in simplifying complex expressions, such as those encountered in calculus.The Pythagorean identity, \( \tan^2 x = \sec^2 x - 1 \), is specifically helpful here. It allows us to simplify expressions involving tangent and secant functions. Using this identity, we're able to switch terms to alternate but equivalent forms, sometimes making integration more straightforward. In the step-by-step solution, this identity was key to transforming the integral into a form where other methods could be applied.

Definite Integrals

Definite integrals are used to calculate the area under the curve of a function over a specified interval. Their notation typically looks like this:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \( a \) and \( b \) are the lower and upper bounds, respectively. However, the exercise we are tackling does not give specific bounds, implying it's an indefinite integral focusing on finding a general solution.Definite integrals would require additional steps beyond the symbolic antiderivative, involving the evaluation of the resulting antiderivative at the specified bounds and then subtracting these values. But remember, every indefinite integral can provide the building blocks you need for evaluating definite integrals effectively.