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**Hint:**Here, we need to find the value of the integral \[\int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}dx}\]. For this, we will use a substitution method. For that, we will make the numerator in such a way that when the denominator or a part of it is differentiated with respect to x, the numerator is obtained. As a result to this, we will get our integral as $\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}dx}$. Then we will substitute tanx as t and when we will differentiate both of them, we will get the value of dx. As a result to this we will obtain our integral in the form of $\int{\dfrac{1}{{{\left( ax+b \right)}^{n}}}dx}$ which will be in the terms of t. then by using the formula $\int{\dfrac{1}{{{\left( ax+b \right)}^{n}}}dx}=\int{{{\left( ax+b \right)}^{-n}}dx}=\dfrac{1}{a}\dfrac{{{\left( ax+b \right)}^{-n+1}}}{-n+1}+C$, we will obtain the required value in terms of t. then we will substitute t back to tanx and hence we will get the required answer.

**Complete step by step answer:**

Here, we need to find the integral \[\int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}dx}\].

For this, we will first try to make it in such a way that when the denominator or a part of it is differentiated with respect to x, the numerator is obtained.

Now, let the integral \[\int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}dx}\] be I.

Hence, we need to find the value of I.

Now, we know that $\dfrac{1}{\cos x}=\sec x$

Putting this in I we get:

$\begin{align}

& \Rightarrow I=\int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}dx} \\

& \Rightarrow I=\int{{{\left( \dfrac{1}{\cos x} \right)}^{2}}\dfrac{1}{{{\left( 1-\tan x \right)}^{2}}}dx} \\

& \Rightarrow I=\int{{{\left( \sec x \right)}^{2}}\dfrac{1}{{{\left( 1-\tan x \right)}^{2}}}dx} \\

& \Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}dx} \\

\end{align}$

Now, we know that when tanx is differentiated with respect to x, we get ${{\sec }^{2}}x$ as a result.

Thus, we will use a substitution method to further solve this integral.

Let tanx=t

Differentiating both sides, we get:

$\begin{align}

& \tan x=t \\

& \Rightarrow {{\sec }^{2}}xdx=dt \\

\end{align}$

Now, putting these values into I, we get:

\[\begin{align}

& I=\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}dx} \\

& I=\int{\dfrac{dt}{{{\left( 1-t \right)}^{2}}}} \\

\end{align}\]

Now, we can see that this integral is in the form of $\int{\dfrac{1}{{{\left( ax+b \right)}^{n}}}dx}$ and we know that this is given as:

$\int{\dfrac{1}{{{\left( ax+b \right)}^{n}}}dx}=\int{{{\left( ax+b \right)}^{-n}}dx}=\dfrac{1}{a}\dfrac{{{\left( ax+b \right)}^{-n+1}}}{-n+1}+C$

Thus, using this in I, we get:

\[I=\int{\dfrac{dt}{{{\left( 1-t \right)}^{2}}}}\]

Here,

n=2

a=-1

b=1

Thus, we get:

\[\begin{align}

& I=\int{\dfrac{dt}{{{\left( 1-t \right)}^{2}}}} \\

& \Rightarrow I=\dfrac{1}{-1}\dfrac{{{\left( 1-t \right)}^{-2+1}}}{-2+1} \\

& \Rightarrow I=-1\left( \dfrac{-1}{1-t} \right) \\

& \Rightarrow I=\dfrac{1}{1-t}+C \\

\end{align}\]

Where, C is the constant of integration.

Now, the integral to solve has been given to us in terms of x, but the answer we obtained is in terms of t. hence, we need to change it back to in terms of x.

We took tanx=t

Hence, t=tanx

Putting it in I, we get:

$\begin{align}

& I=\dfrac{1}{1-t}+C \\

& \Rightarrow I=\dfrac{1}{1-\tan x}+C \\

\end{align}$

**Hence, the required answer is $\dfrac{1}{1-\tan x}+C$.**

**Note:**Remember that there are more than one ways most of the time to solve the integrals specially the ones which have trigonometry involved which results. And since the integration done here is indefinite, it may result in two different answers. But we should not panic at that instant because there can be two different answers to one indefinite integration question because of the constant of integration present. It can be any value and hence can change our answer into a different form resulting in a different answer. But when it comes to definite integration, whatever may be the integral, after putting the limits the answer is always the same and unique.

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