Question

\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]

Answer 1

\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right)dx\]
\[ = \int x \left( \frac{\frac{1}{\cos 2x} - 1}{\frac{1}{\cos 2x} + 1} \right)dx\]
\[ = \int x \left( \frac{1 - \cos 2x}{1 + \cos 2x} \right)dx\]
\[ = \int x \left( \frac{2 \sin^2 x}{2 \cos^2 x} \right)dx \left[ \because \left( 1 - \cos 2x \right) = 2 \sin^2 x and \left( 1 + \cos 2x \right) = 2 \cos^2 x \right]\]
\[ = \int x . \tan^2 \text{ x dx  }\]
\[ = \int x . \left( \sec^2 x - 1 \right) dx\]
\[ = \int x_I . \sec^2_{II}   \text{ x   dx } - \int \text{ x dx }\]
\[ = x\int \sec^2\text{  x dx }- \int\left\{ \frac{d}{dx}\left( x \right)\int\sec^2  \text{ x  dx }\right\}dx - \frac{x^2}{2} + C_1 \]
\[ = x \tan x - \int1 . \text{ tan x dx } - \frac{x^2}{2} + C_1 \]
\[ = x \tan x - \text{ log  }\left| \sec x \right| - \frac{x^2}{2} + C_2 + C_1 \]
\[ = x \tan x - \text{ log }\left| \sec x \right| - \frac{x^2}{2} + C \left( \text{ where C} = C_1 + C_2 \right)\]