Evaluate the integral. \( \int 23 \tan ^{5} x \sec ^{4} x d x \) \( \int 23 \tan ^{5} x \sec ^{4} x d x=\square \)

To evaluate the integral \( \int 23 \tan ^{5} x \sec ^{4} x \, dx \), we can use substitution. Let \( u = \tan x \), then \( du = \sec^2 x \, dx \). We need to express \( \sec^4 x \) in terms of \( u \). Since \( \sec^2 x = 1 + \tan^2 x = 1 + u^2 \), we get: \[ \sec^4 x = (1 + u^2)^2 \] Now, substituting into the integral, we also need to fix the \( dx \) term: \[ dx = \frac{du}{\sec^2 x} = \frac{du}{1 + u^2} \] The integral becomes: \[ \int 23 u^5 (1 + u^2)^2 \frac{du}{1 + u^2} = \int 23 u^5 (1 + u^2) \, du \] Now we can expand and integrate: \[ \int 23 (u^5 + u^7) \, du = 23 \left( \frac{u^6}{6} + \frac{u^8}{8} \right) + C \] Substituting back \( u = \tan x \): \[ = \frac{23}{6} \tan^6 x + \frac{23}{8} \tan^8 x + C \] Thus, the evaluated integral is: \[ \int 23 \tan ^{5} x \sec ^{4} x \, dx = \frac{23}{6} \tan^6 x + \frac{23}{8} \tan^8 x + C \]