STD 12 - 7.1 inderfinite integral — JEE कक्षा 11 साइन्स — Question

$ = \int_{}^{} {x\tan x\,.\,{{\sec }^2}x\,dx} $

$\tan x = t $ रखने पर $  {\sec ^2}x\,dx = dt$ ओर  $x = {\tan ^{ - 1}}t,$ 

$\int_{}^{} {{{\tan }^{ - 1}}t\,.\,t\,dt} = \frac{{x{{\tan }^2}x}}{2} - \frac{1}{2}t + \frac{1}{2}{\tan ^{ - 1}}t$

$ = \frac{{x({{\sec }^2}x - 1)}}{2} - \frac{1}{2}\tan x + \frac{1}{2}x = \frac{1}{2}[x{\sec ^2}x - \tan x] + c$.