1. We are asked to evaluate the integral $\int \tan^2 x \sec^4 x \, dx$. 2. Recall the identities: - $\sec^2 x = 1 + \tan^2 x$ - $\frac{d}{dx}(\tan x) = \sec^2 x$ 3. Rewrite $\sec^4 x = (\sec^2 x)^2$ and express in terms of $\tan x$: $$\int \tan^2 x \sec^4 x \, dx = \int \tan^2 x (\sec^2 x)^2 \, dx$$ 4. Use substitution $u = \tan x$, so $du = \sec^2 x \, dx$. 5. Then: $$\int \tan^2 x \sec^4 x \, dx = \int u^2 (\sec^2 x)^2 \, dx = \int u^2 \sec^2 x (\sec^2 x \, dx) = \int u^2 \sec^2 x \, du$$ 6. But $\sec^2 x = 1 + u^2$, so: $$\int u^2 (1 + u^2) \, du = \int (u^2 + u^4) \, du$$ 7. Integrate term by term: $$\int u^2 \, du + \int u^4 \, du = \frac{u^3}{3} + \frac{u^5}{5} + C$$ 8. Substitute back $u = \tan x$: $$\boxed{\frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C}$$