$\cfrac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}$
$\cfrac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
: Let $I = \int {\cfrac{1}{{{{\cos }^2}x{{\left( {1 - \tan x} \right)}^2}}}dx = I = \int {\cfrac{{{{\sec }^2}x}}{{{{\left( {1 - \tan x} \right)}^2}}}dx} }$
Put $1 - \tan x = t$ $\Rightarrow$ $- {\sec ^2}xdx = dt$
$\therefore$ $I = - \int {\cfrac{{dt}}{{{t^2}}}} = - \cfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \cfrac{1}{t} + C = \cfrac{1}{{1 - \tan x}} + C$
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