Integrals — Class 12 Maths Solution

.: Let $I = \int {\cfrac{1}{{\sin x{{\cos }^3}x}}dx} = \int {\cfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x{{\cos }^3}x}}} dx$

$= \int {\left( {\cfrac{{{{\sin }^2}x}}{{\sin x{{\cos }^3}x}} + \cfrac{{{{\cos }^2}x}}{{\sin x{{\cos }^3}x}}} \right)dx} = \int {\left( {\cfrac{{\sin x}}{{{{\cos }^3}x}} + \cfrac{{\cos x}}{{\sin x{{\cos }^2}x}}} \right)} dx$

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

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

$\therefore$ $I = \int {\left( {t + \cfrac{1}{t}} \right)} dt = \cfrac{{{t^2}}}{2} + \log \left| t \right| + C = \log \left| {\tan x} \right| + \cfrac{1}{2}{\tan ^2}x + C$