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Integrate : $\int x\bigg( \large\frac{\sec 2x -1}{\sec 2x+1}\bigg) dx$

$(a)\;x \tan x +\log |\sec x | -\frac{x^2}{2}+c \\(b)\;x \tan x -\log |\sec x | -\frac{x^2}{2}+c \\(c)\; -x \tan x -\log |\sec x | -\frac{x^2}{2}+c \\ (d)\;none $

1 Answer

$\int x \bigg( \large\frac{\sec 2x-1}{\sec 2 x +1}\bigg)$$dx$
$\int x \bigg( \large\frac{1-\cos 2x}{1+ \cos 2x}\bigg)$$dx$
=> $\int x. \tan ^2 x dx$
=> $\int x(\sec^2 x -1)dx$
=> $\int x \sec^2 x dx -\int x dx$
=> $x. \tan x -\int \tan x dx -\large\frac{x^2}{2}$
=> $ x \tan x - \log | \sec x| -\large\frac{x^2}{2}$$+c$
Hence b is the correct answer.


answered Jan 2, 2014 by meena.p
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