Question

Evaluate the following:

${\displaystyle \int {\tan }^{2}x{\sec }^{4}x\text{d}x}$

Answer 1

Let I = ${\displaystyle \int {\tan }^{2}x{\sec }^{4}x\text{d}x}$

= ${\displaystyle \int {\tan }^{2}x{\sec }^{2}x{\sec }^{2}x\text{d}x}$

= ${\displaystyle \int {\tan }^{2}x(1+{\tan }^{2}x){\sec }^{2}x\text{d}x}$

Put tan x = t

⇒ ${\displaystyle {\sec }^{2}x\text{d}x}$ = dt

∴ I = ${\displaystyle \int {\text{t}}^2(1+{\text{t}}^2)\text{dt}}$

= ${\displaystyle \int ({\text{t}}^2+{\text{t}}^4)\text{dt}}$

= ${\displaystyle \frac{{\text{t}}^3}{3}+\frac{{\text{t}}^5}{5}+\text{C}}$

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