Question

Find the integral. Use a computer algebra system to confirm your result. $$ \int\left(\tan ^{4} t-\sec ^{4} t\right) d t $$

Short answer

The solution is \( I = \int\left(\tan ^{4} t-\sec ^{4} t\right) dt = \frac{1}{3}t \tan t - \frac{2}{3} \ln |\sec| + t - \frac{1}{3} \tan t (\tan^2 t - 2) - \frac{2}{3} \tan t \)

Step-by-step solution

Simplification

Separate the given integral into two parts. So, we get two integrals: \[ \int \tan^{4} t \, dt \] and \[ \int \sec^{4} t \, dt \]

Calculating the first integral

To solve the first integral, we will use the identity \(\tan^2 t = \sec^2 t - 1\) and write \(\tan^4 t\) as \((\sec^2 t - 1)^2\). The integral then becomes: \[ \int (\sec^2 t - 1)^2 dt \] Now, we can further simplify the integral by expanding it: \[ \int (\sec^4 t - 2\sec^2 t + 1) dt \] The answer for this integral, let's denote it with \( I_1 \), is \[ I_1 = \frac{1}{3}t \tan t - \frac{2}{3} \ln |\sec| + t \]

Calculating the second integral

The second integral \(\int \sec^{4} t dt\) is a standard integral. Using the formula for the integral of \(\sec^n t\), where \(n\) is even and greater than 1, the integral can be written in terms of \(\tan t\) and a lower powered \(\sec t\). Thus, \(\int \sec^{4} t dt\) becomes \( \frac{1}{3}\tan t(\sec^2 t - 2) + \frac{2}{3}\int \sec^2 t dt \). Using the integral \(\int \sec^2 t dt = \tan t\), the calculation of the second integral \(I_2\) results in \[ I_2 = \frac{1}{3} \tan t (\tan^2 t - 2) + \frac{2}{3} \tan t \]

Putting it all together

Subtract \(I_2\) from \(I_1\) to get the final answer for the original integral. Therefore, the solution for the given integral is: \[ I = \int\left(\tan ^{4} t-\sec ^{4} t\right) dt = I_1- I_2 = \(\frac{1}{3}t \tan t - \frac{2}{3} \ln |\sec| + t - \frac{1}{3} \tan t (\tan^2 t - 2) - \frac{2}{3} \tan t \) \]