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 Q25  Integrate the functions \frac{1 }{ \cos ^2 x (1-\tan x )^2}

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Given function  \frac{1 }{ \cos ^2 x (1-\tan x )^2},

or simplified as \frac{1 }{ \cos ^2 x (1-\tan x )^2} = \frac{\sec^2x}{(1-\tan x)^2}

Assume the (1-\tan x)=t

 \therefore -\sec^2xdx =dt

\implies \int \frac{\sec^2x}{(1-\tan x)^2}dx = \int\frac{-dt}{t^2}

= -\int t^{-2} dt

= \frac{1}{t} +C

Now, back substituted the value of t.

= \frac{1}{1-\tan x}+C

 where C is any constant value.

Posted by

Divya Prakash Singh

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