Question

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

Short 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-by-step solution

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:\(-\int \sec^{4}(u) \tan(u) du\).

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.

Substitute u back

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

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.