A) \[\frac{1}{\tan x-1}+c\]
B) \[\frac{1}{1-\tan x}+c\]
C) \[-\frac{1}{3}\frac{1}{{{(1-\tan x)}^{3}}}+c\]
D) None of these
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{1}{{{\cos }^{2}}x{{(1-\tan x)}^{2}}}dx=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{(\tan x-1)}^{2}}}}}\] Put \[\tan x-1=t\Rightarrow {{\sec }^{2}}x\,dx=dt,\] then it reduces to \[\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt}=\frac{-1}{\tan x-1}+c=\frac{1}{1-\tan x}+c.\]
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