Consider the expression\[\int{{{\sec }^{n}}x\tan xdx}\]. Find the value of the integral.

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Hint: You can rewrite \[{{\sec }^{n}}x\]as \[{{\sec }^{n-1}}x.\sec x\], in the given integral. Later, you can employ the substitution method to compute the given integral, by substituting\[\sec x=t\].

We must evaluate the integral of \[\int{{{\sec }^{n}}x\tan xdx}\].
Let us assume the given integral as \[\int{{{\sec }^{n}}x\tan xdx}=I\].
We can rewrite \[{{\sec }^{n}}x\]as \[{{\sec }^{n-1}}x.\sec x\], since we know \[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
Therefore, the integer can be expressed as,
\[I=\int{{{\sec }^{n-1}}x.\sec x.tanxdx}\].
Let us use the substitution process for evaluating this particular form of integral.
So, let us put \[\sec x\]as ‘t’.
\[\sec x=t\].
Differentiating on both the sides of the above equation, we have:
\[\dfrac{d\left( \sec x \right)}{dx}=\dfrac{dt}{dx}\]
\[\sec x.\tan x=\dfrac{dt}{dx}\]
Therefore, \[dt=\sec x.\tan x.dx\]
As we substitute the value of ‘t’ and \[dt\]in the integrals I, the integral will transform as mentioned below:
\[I=\int{{{\left( t \right)}^{n-1}}dt}\]
Evaluating the integral further, we have:
\[I=\dfrac{{{t}^{\left( n-1 \right)+1}}}{\left( n-1 \right)+1}+c\].
Since, for any given x,\[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}\]
Therefore, the integral reduces to:
As, we have \[t=\sec x\], let us put it back in the solved expression of the integral.
Then, we have:
\[I=\dfrac{{{\sec }^{n}}x}{n}+c\]
Where, c is any constant.
So, by following the process of substitution we have evaluated the given integral.
Hence the answer for the given integral is \[\dfrac{{{\sec }^{n}}x}{n}+c.\]

Note: We can directly evaluate the given integral by using the formula \[\int{{{\left( f\left( x \right) \right)}^{n}}{f}'\left( x \right)dx=\dfrac{f{{\left( x \right)}^{n+1}}}{n+1}+c}\] where \[f\left( x \right)={{\sec }^{n-1}}\]and \[{f}'\left( x \right)=\sec x\tan x\] respectively. Using shortcut methods effectively will save time and give a smart approach to the answer. Also, applying product rule is not recommended in this case as the process will be very lengthy and difficult to solve.
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