Evaluate $\int \tan^2 x \sec^4 x dx$.

$\frac{\tan^5 x}{5} + \frac{\tan^3 x}{3} + C$

The correct answer is Option (2) → $\frac{\tan^5 x}{5} + \frac{\tan^3 x}{3} + C$

Let $I = \int \tan^2 x \sec^4 x dx$

$= \int \tan^2 x \cdot \sec^2 x \cdot \sec^2 x dx$

$= \int \tan^2 x (1 + \tan^2 x) \cdot \sec^2 x dx$

Put $\tan x = t \Rightarrow \sec^2 x dx = dt$

$∴I = \int t^2(1 + t^2) dt = \int (t^2 + t^4) dt$

$= \frac{t^3}{3} + \frac{t^5}{5} + C = \frac{\tan^5 x}{5} + \frac{\tan^3 x}{3} + C$