${\tan ^4}x$

: Let$I = \int {{{\tan }^4}x} dx = \int {{{\left( {{{\sec }^2}x - 1} \right)}^2}dx}$

$= \int {\left( {{{\sec }^4}x - 2{{\sec }^2}x + 1} \right)} dx = \int {{{\sec }^4}x} dx - 2\int {{{\sec }^2}x} dx + \int {\left( 1 \right)dx}$

$= \int {{{\sec }^4}xdx - 2\tan x + x + {C_1}}$ $\Rightarrow$ $I = {I_1} - 2\tan x + x + {C_1}$

…(i)
Where ${I_1} = \int {{{\sec }^4}x} dx$
Now, ${I_1} = \int {{{\sec }^4}xdx} = \int {{{\sec }^2}x \cdot {{\sec }^2}xdx}$

$= \int {\left( {1 + {{\tan }^2}x} \right)} {\sec ^2}x\,dx$

Put $\tan x = t$ $\Rightarrow$ ${\sec ^2}xdx = dt$

$\therefore$ ${I_1} = \int {\left( {1 + {t^2}} \right)dt} = t + \cfrac{{{t^3}}}{3} + {C_2} = \tan x + \cfrac{1}{3}{\tan ^3}x + {C_2}$

….(ii)
From (i) and (ii),

we have
$I = \tan x + \cfrac{1}{3}{\tan ^3}x + {C_2} - 2\tan x + x + {C_1}$

$= \cfrac{1}{3}{\tan ^3}x - \tan x + x + C$ ,where $C = {C_1} + {C_2}$