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$\int {{{\sec }^n}x\tan xdx} $

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Last updated date: 13th Jun 2024
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Answer
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Hint: This problem deals with integrations. There are two types of integrals which are definite integrals and indefinite integrals. Indefinite integrals have no limits on the integral unlike the definite integrals. There are three methods to solve the indefinite integrals. The first method is integration by substitution. Whereas the second method is integration by partial fractions. The third method is integration by parts.
In this problem we are going to do the integration of the given integral by the method of integration by substitution.

Complete step-by-step answer:
Integration by substitution is the method where the function inside the integral is assigned another variable of integration. That is we are going to substitute another variable instead of the original function, which makes the integration process much easier.
Given an indefinite integral, we have to find the integral of $\int {{{\sec }^n}x\tan xdx} $.
Consider the integral $\int {{{\sec }^n}x\tan xdx} $, as given below:
$ \Rightarrow \int {{{\sec }^n}x\tan xdx} $
$ \Rightarrow \int {{{\sec }^{n - 1 + 1}}x\tan xdx} $
Here ${\sec ^{n - 1 + 1}}x = {\sec ^{n - 1}}x\sec x$, as expressed below in the integral.
$ \Rightarrow \int {{{\sec }^{n - 1}}x\sec x\tan xdx} $
Applying the integration by substitution method on the integral, as given below:
Let $\sec x = t$ ;
Now differentiate the above equation on both sides, as given below:
$ \Rightarrow \sec x\tan xdx = dt$
Here $\sec x = t$, hence ${\sec ^{n - 1}}x = {t^{n - 1}}$
Now substitute all the above expressions in the integral $\int {{{\sec }^{n - 1}}x\sec x\tan xdx} $, as given below:
$ \Rightarrow \int {{t^{n - 1}}dt} $
$ \Rightarrow \dfrac{{{t^{n - 1 + 1}}}}{{n - 1 + 1}} + c$
$ \Rightarrow \dfrac{{{t^n}}}{n} + c$
Substitute back what the variable $t$ was assigned for $t = \sec x$, as given below:
$ \Rightarrow \dfrac{{{{\sec }^n}x}}{n} + c$
$\therefore \int {{{\sec }^n}x\tan xdx} = \dfrac{{{{\sec }^n}x}}{n} + c$

$\int {{{\sec }^n}x\tan xdx} = \dfrac{{{{\sec }^n}x}}{n} + c$

Note:
Please note that this problem of integration can also be done by the method of integration by parts, which is the integration of the product of two functions which is given by the formula of integration by parts.
Similarly applying this formula to the given integral $\int {{{\sec }^n}x\tan xdx} $, but here the ${f_1}(x) = {\sec ^{n - 2}}x$ and ${f_2}(x) = {\sec ^2}x\tan x$, and proceeding by substitution for $\tan x = t$. Either of the methods give the same final answer.