Integrals — Class 12 Maths Solution

: Let $I = \int {{{\tan }^2}\left( {2x - 3} \right)dx} = \int {\left[ {{{\sec }^2}\left( {2x - 2} \right) - 1} \right]dx}$

$= \int {{{\sec }^2}\left( {2x - 3} \right)dx} - \int {dx} = \int {{{\sec }^2}\left( {2x - 3} \right)} dx - x + {C_1}$

$= {I_1} - x + {C_1}$

….(i)
Where ${I_1} = \int {{{\sec }^2}\left( {2x - 3} \right)dx}$

Put $2x - 3 = t$ $\Rightarrow$ $2dx = dt$
$\Rightarrow$ ${I_1} = \cfrac{1}{2}\int {{{\sec }^2}t} dt = \cfrac{1}{2}\tan t + {C_2} = \cfrac{1}{2}\tan \left( {2x - 3} \right) + {C_2}$

…..(ii)
From (i) and (ii) ,

we get
$I = {I_1} - x + {C_1} = \cfrac{1}{2}\tan \left( {2x - 3} \right) - x + C$ where $C = {C_1} + {C_2}$