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

Last updated date: 24th Jul 2024
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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.