Problem 60 - Essential Calculus

Question

Use a computer algebra system to find the integral. Graph the antiderivatives for two different values of the constant of integration.$$ \int \sec ^{4}(1-x) \tan (1-x) d x $$

Step-by-Step Solution

Answer

The integral of \( \sec^{4}(1-x) \tan(1-x) \) with respect to x is \(-\sec^{2}(1-x) \sec(1-x) + \sec(1-x) \tan(1-x) + ln|\sec(1-x) + \tan(1-x)| + C.\) Wifferent constants of integration will shift the graph of the antiderivative vertically.

Step 1: Recognize the pattern and use substitution

Note a pattern in our integrand. We know that the derivative of \(\sec(u)\) is \(\sec(u)\tan(u)\). Seeing that we have a function of this form in our integral and we have an extra \(\sec^{2}(1-x)\) factor that can be absorbed in \(du\) substitution, we can set \(u = 1 - x\), the argument of sec and tan functions. So, \(du = -dx\). We replace these in our integral, and remember to change the sign because \(du = -dx\). This changes our integral to:\(-\int \sec^{4}(u) \tan(u) du\).

Step 2: Integrating with respect to u

Next, rewrite the integral as a product of two secant functions and integrate by parts:\[-\int \sec^{2}(u) \cdot \sec^{2}(u) \tan(u) du.\]Set \(v = \sec^{2}(u)\) and \(dw = \sec^{2}(u) \tan(u) du\). Then, \(dv = 2 \sec(u) \tan(u)\) sec(u) du and \(w = \sec(u)\). Now, apply the integration by parts formula: \[\int v dw = vw - \int w dv.\] Substitute the values we just calculated: \[-\sec^{2}(u) \sec(u) + 2 \int \sec^{3}(u) du.\]Integrating \(\sec^{3}(u)\) can be tricky, so you may need to look that up and that integral is \(\frac{1}{2}\) sec(u) tan(u) + \(\frac{1}{2}\)ln|sec(u) + tan(u)|.So, our integral becomes: \[-\sec^{2}(u) \sec(u) + \sec(u) tan(u) + ln|sec(u) + tan(u)| + C,\] where C is the constant of integration.

Step 3: Substitute u back

After the integral is computed, now substitute \(u = 1 - x\) back into the equation: \[-\sec^{2}(1-x) \sec(1-x) + \sec(1-x) \tan(1-x) + ln|\sec(1-x) + \tan(1-x)| + C.\]

Step 4: Plot the antiderivative

Plot this function \(-\sec^{2}(1-x) \sec(1-x) + \sec(1-x) \tan(1-x) + ln|\sec(1-x) + tan(1-x)| + C\) for two different values of C, say 0 and 1. The resulting graphs will be vertically shifted up or down by the constant of integration.

Key Concepts

Definite Integrals

Definite integrals are a vital part of calculus and mathematics in general. They help us find the area under a curve between two points on a graph.
For the function \(f(x)\), the definite integral over the interval \([a, b]\) can be written as:

• \( \int_{a}^{b} f(x) \, dx \)

This gives us a specific numerical value.
Unlike indefinite integrals, which include a constant \(C\), definite integrals are calculated over a specific range and do not include \(C\).
They provide real-world applications in areas like physics, engineering, and economics.

Antiderivative

An antiderivative is a function whose derivative gives back the original function.
If \(F(x)\) is an antiderivative of \(f(x)\), then:

• \( F'(x) = f(x) \)

When solving integrals, finding the antiderivative is key, as it helps determine the indefinite integral.
Remember that the general form of an antiderivative includes an arbitrary constant \(C\).
This recognizes the fact that differentiating any constant gives zero. For instance, \(\int x^2 \, dx = \frac{1}{3}x^3 + C\) because differentiating \(\frac{1}{3}x^3 + C\) gives \(x^2\).

Integration by Parts

Integration by parts is a useful technique derived from the product rule for differentiation.
It's often used when the integrand is a product of two functions. The formula is:

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

Here, you choose parts of the integrand to be \(u\) and \(dv\), then differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\).
This strategy transforms a complicated integral into a hopefully simpler one.
In the given problem, integration by parts helps us manage complex secant terms and find the integral step-by-step, breaking it into easier pieces.

Substitution Method

The substitution method, also known as "u-substitution," simplifies integrals by changing variables.
This technique is especially helpful when an integral contains a composite function.
In the substitution method:

• Choose a substitution \(u = g(x)\)
• Find \(du = g'(x) \, dx\)
• Rewrite the integral in terms of \(u\)

After integrating with respect to \(u\), substitute back the expression for \(x\).
This method streamlines problem-solving with complex integrands, as demonstrated in the provided example
where setting \(u = 1-x\) led to a more straightforward problem.