1. **Problem statement:** Evaluate the integral $\int \tan^2 x \sec^4 x \, dx$. 2. **Recall formulas and identities:** - $\sec^2 x = 1 + \tan^2 x$ - Derivative: $\frac{d}{dx}(\tan x) = \sec^2 x$ 3. **Rewrite the integral:** $$\int \tan^2 x \sec^4 x \, dx = \int \tan^2 x (\sec^2 x)^2 \, dx$$ 4. **Use substitution:** Let $u = \tan x$, then $du = \sec^2 x \, dx$. 5. **Express integral in terms of $u$:** $$\int \tan^2 x \sec^4 x \, dx = \int u^2 (\sec^2 x)^2 \, dx = \int u^2 \sec^4 x \, dx$$ Since $du = \sec^2 x \, dx$, then $\sec^2 x \, dx = du$, so $\sec^4 x \, dx = \sec^2 x (\sec^2 x \, dx) = \sec^2 x du$. Rewrite integral as: $$\int u^2 \sec^2 x \, du$$ 6. **Rewrite $\sec^2 x$ in terms of $u$:** Since $u = \tan x$, $\sec^2 x = 1 + u^2$. 7. **Substitute:** $$\int u^2 (1 + u^2) \, du = \int (u^2 + u^4) \, du$$ 8. **Integrate term-by-term:** $$\int u^2 \, du + \int u^4 \, du = \frac{u^3}{3} + \frac{u^5}{5} + C$$ 9. **Back-substitute $u = \tan x$:** $$\frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$$ **Final answer:** $$\int \tan^2 x \sec^4 x \, dx = \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$$