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

Answer

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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:

\[I=\dfrac{{{t}^{n}}}{n}+c\]

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.

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:

\[I=\dfrac{{{t}^{n}}}{n}+c\]

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.

Last updated date: 26th May 2023

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