$\int \sqrt[7]{\tan (8 x)} \sec ^{2}(8 x) d x$

Step 1 ：Let \(u = \tan(8x)\). This implies that \(du = 8\sec^2(8x)dx\) or \(dx = \frac{1}{8\sec^2(8x)}du\).

Step 2 ：Substitute \(u\) and \(dx\) into the integral to get \(\frac{1}{8}\int u^{1/7} du\).

Step 3 ：Evaluate the integral to get \(\frac{7}{8}u^{8/7} + C\), where \(C\) is the constant of integration.

Step 4 ：Substitute \(u = \tan(8x)\) back into the result to get \(\frac{7}{8}\tan^{8/7}(8x) + C\).

Step 5 ：Differentiate \(\frac{7}{8}\tan^{8/7}(8x) + C\) with respect to \(x\) to check the result. The derivative is indeed \(\sqrt[7]{\tan (8 x)} \sec ^{2}(8 x)\), so the result is correct.

Step 6 ：\(\boxed{\frac{7}{8}\tan^{8/7}(8x) + C}\) is the final answer.