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Question

Question: \(\int {{{\sec }^n}x\tan xdx} \)...

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

Explanation

Solution

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 secnxtanxdx\int {{{\sec }^n}x\tan xdx} .
Consider the integral secnxtanxdx\int {{{\sec }^n}x\tan xdx} , as given below:
secnxtanxdx\Rightarrow \int {{{\sec }^n}x\tan xdx}
secn1+1xtanxdx\Rightarrow \int {{{\sec }^{n - 1 + 1}}x\tan xdx}
Here secn1+1x=secn1xsecx{\sec ^{n - 1 + 1}}x = {\sec ^{n - 1}}x\sec x, as expressed below in the integral.
secn1xsecxtanxdx\Rightarrow \int {{{\sec }^{n - 1}}x\sec x\tan xdx}
Applying the integration by substitution method on the integral, as given below:
Let secx=t\sec x = t ;
Now differentiate the above equation on both sides, as given below:
secxtanxdx=dt\Rightarrow \sec x\tan xdx = dt
Here secx=t\sec x = t, hence secn1x=tn1{\sec ^{n - 1}}x = {t^{n - 1}}
Now substitute all the above expressions in the integral secn1xsecxtanxdx\int {{{\sec }^{n - 1}}x\sec x\tan xdx} , as given below:
tn1dt\Rightarrow \int {{t^{n - 1}}dt}
tn1+1n1+1+c\Rightarrow \dfrac{{{t^{n - 1 + 1}}}}{{n - 1 + 1}} + c
tnn+c\Rightarrow \dfrac{{{t^n}}}{n} + c
Substitute back what the variable tt was assigned for t=secxt = \sec x, as given below:
secnxn+c\Rightarrow \dfrac{{{{\sec }^n}x}}{n} + c
secnxtanxdx=secnxn+c\therefore \int {{{\sec }^n}x\tan xdx} = \dfrac{{{{\sec }^n}x}}{n} + c

secnxtanxdx=secnxn+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 secnxtanxdx\int {{{\sec }^n}x\tan xdx} , but here the f1(x)=secn2x{f_1}(x) = {\sec ^{n - 2}}x and f2(x)=sec2xtanx{f_2}(x) = {\sec ^2}x\tan x, and proceeding by substitution for tanx=t\tan x = t. Either of the methods give the same final answer.