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Hint: We know that integration of anything is the reverse process of differentiation. In this question, we will solve this by using some basic integration and basic differentiation. You can start your solution by writing sec x is equal to $\dfrac{1}{\cos x}$ and tan x is equal to $\dfrac{\sin x}{\cos \ x}$.

Complete step-by-step answer:

It is given in the question that to integrate sec x. tan x

\[\int{\tan x.\sec xdx........................\left( i \right)}\]

Here, in equation (i) we can write tan x as $\dfrac{\sin x}{\cos \ x}$ and sec x as $\dfrac{1}{\cos x}$.

\[\begin{align}

& =\int{\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x}}dx \\

& =\int{\dfrac{\sin x}{{{\left( \cos x \right)}^{2}}}dx.........................\left( ii \right)} \\

\end{align}\]

Let us assume that u = cos x.

Then differentiating u = cos x with respect to x; we get,

$\dfrac{du}{dx}=-\sin x$

We can write du = -sin x . dx

So, putting -du in place of sin x dx and ${{\left( u \right)}^{2}}$ in place of ${{\left( \cos x \right)}^{2}}$ in equation (ii), we get;

$=-\int{\dfrac{du}{{{\left( u \right)}^{2}}}.................\left( iii \right)}$

We can write $\dfrac{1}{{{\left( u \right)}^{2}}}\ as{{\left( u \right)}^{-2}}$ in equation (iii) we get,

$=-\int{{{\left( u \right)}^{-2}}du}$

Now, we have to integrate \[-{{\left( u \right)}^{-2}}\] with respect to u and we know the basic integration formula;

$\int{{{x}^{n}}dn}=\dfrac{{{x}^{n+1}}}{n+1}+c$

So, on integrating \[-{{\left( u \right)}^{-2}}\]with respect to u we get;

$\begin{align}

& =\dfrac{-{{\left( u \right)}^{-2+1}}}{-2+1}+c \\

& =\dfrac{-{{u}^{-1}}}{-1}+c \\

& ={{u}^{-1}}+c \\

& =\dfrac{1}{u}+c....................\left( iv \right) \\

\end{align}$

Also, we had assumed earlier that u = cos x. So, putting the value of u = cos x in equation (iv), we get;

$=\dfrac{1}{\cos x}+c.............\left( v \right)$

We know that $\dfrac{1}{\cos x}=\sec x$so, writing \[\dfrac{1}{\cos x}\ as\ \sec x\] in equation (v), we get = sec x + c.

Thus, we get $\int{\sec x.\tan x=\sec x+c.}$

Note: Integration of anything is the reverse of the process of differentiation of something. So if you know the correct answer but if you have any confusion in it then you can solve this question by this alternative method.

As in this question, we have to find $\int{\tan x.\sec x}$ with respect to x. So, we will differentiate sec x with respect to x to get $\int{\tan x.\sec x}$.

So, we have;

$u=\sec x\ or\ u=\dfrac{1}{\cos x}$ on differentiating with respect to x, we get;

\[\begin{align}

& \dfrac{du}{dx}=-\dfrac{1.\left( -\sin x \right)}{{{\cos }^{2}}x} \\

& \dfrac{du}{dx}=\dfrac{\sin x}{{{\cos }^{2}}x} \\

& \dfrac{du}{dx}=\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x} \\

& \dfrac{du}{dx}=\sec x.\tan x \\

& du=\sec x.\tan x.dx \\

& \dfrac{d\left( \sec x \right)}{dx}=\sec x.\tan x \\

& or \\

& \int{\sec x\tan x.dx=\sec x}. \\

\end{align}\]

Complete step-by-step answer:

It is given in the question that to integrate sec x. tan x

\[\int{\tan x.\sec xdx........................\left( i \right)}\]

Here, in equation (i) we can write tan x as $\dfrac{\sin x}{\cos \ x}$ and sec x as $\dfrac{1}{\cos x}$.

\[\begin{align}

& =\int{\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x}}dx \\

& =\int{\dfrac{\sin x}{{{\left( \cos x \right)}^{2}}}dx.........................\left( ii \right)} \\

\end{align}\]

Let us assume that u = cos x.

Then differentiating u = cos x with respect to x; we get,

$\dfrac{du}{dx}=-\sin x$

We can write du = -sin x . dx

So, putting -du in place of sin x dx and ${{\left( u \right)}^{2}}$ in place of ${{\left( \cos x \right)}^{2}}$ in equation (ii), we get;

$=-\int{\dfrac{du}{{{\left( u \right)}^{2}}}.................\left( iii \right)}$

We can write $\dfrac{1}{{{\left( u \right)}^{2}}}\ as{{\left( u \right)}^{-2}}$ in equation (iii) we get,

$=-\int{{{\left( u \right)}^{-2}}du}$

Now, we have to integrate \[-{{\left( u \right)}^{-2}}\] with respect to u and we know the basic integration formula;

$\int{{{x}^{n}}dn}=\dfrac{{{x}^{n+1}}}{n+1}+c$

So, on integrating \[-{{\left( u \right)}^{-2}}\]with respect to u we get;

$\begin{align}

& =\dfrac{-{{\left( u \right)}^{-2+1}}}{-2+1}+c \\

& =\dfrac{-{{u}^{-1}}}{-1}+c \\

& ={{u}^{-1}}+c \\

& =\dfrac{1}{u}+c....................\left( iv \right) \\

\end{align}$

Also, we had assumed earlier that u = cos x. So, putting the value of u = cos x in equation (iv), we get;

$=\dfrac{1}{\cos x}+c.............\left( v \right)$

We know that $\dfrac{1}{\cos x}=\sec x$so, writing \[\dfrac{1}{\cos x}\ as\ \sec x\] in equation (v), we get = sec x + c.

Thus, we get $\int{\sec x.\tan x=\sec x+c.}$

Note: Integration of anything is the reverse of the process of differentiation of something. So if you know the correct answer but if you have any confusion in it then you can solve this question by this alternative method.

As in this question, we have to find $\int{\tan x.\sec x}$ with respect to x. So, we will differentiate sec x with respect to x to get $\int{\tan x.\sec x}$.

So, we have;

$u=\sec x\ or\ u=\dfrac{1}{\cos x}$ on differentiating with respect to x, we get;

\[\begin{align}

& \dfrac{du}{dx}=-\dfrac{1.\left( -\sin x \right)}{{{\cos }^{2}}x} \\

& \dfrac{du}{dx}=\dfrac{\sin x}{{{\cos }^{2}}x} \\

& \dfrac{du}{dx}=\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x} \\

& \dfrac{du}{dx}=\sec x.\tan x \\

& du=\sec x.\tan x.dx \\

& \dfrac{d\left( \sec x \right)}{dx}=\sec x.\tan x \\

& or \\

& \int{\sec x\tan x.dx=\sec x}. \\

\end{align}\]

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