Question

To evaluate the following integral,
$\int{{{\sec }^{6}}x\tan xdx}$.

Answer 1

Hint: For solving this problem, one needs to be aware about the concept of indefinite integration. Further, for solving questions where multiple terms are present as product terms inside the integration, we can use the technique of substitution.

Complete step by step answer:
Before solving the problem, we try to understand some basics about indefinite integration. Integration is the reverse process of differentiation. Thus, it is basically an anti-derivative process, thus to explain. We know that-
$\dfrac{d(\sin x)}{dx}=\cos x$
Thus, since integration is the reverse process of differentiation, thus $\int{\cos xdx}$= sinx.
Now, we do a similar process for the above problem. Since, we have,
$\int{{{\sec }^{6}}x\tan xdx}$ -- (A)
We use the technique of substitution. Thus, let’s substitute t=tanx. Now, $\dfrac{dt}{dx}$= ${{\sec }^{2}}x$-- (1)
(Since, $\dfrac{d(\tan x)}{dx}={{\sec }^{2}}x$)
From (1), we have,
dt = ${{\sec }^{2}}x$dx
dx = $\dfrac{dt}{{{\sec }^{2}}x}$
We substitute this in (A), we get,
$\int{({{\sec }^{6}}x)(\operatorname{t})\left( \dfrac{dt}{{{\sec }^{2}}x} \right)}$
Since, t = tanx. Thus,
$\int{({{\sec }^{4}}x)(\operatorname{t})dt}$
Now, we know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$, thus,
$\begin{aligned} & \int{{{(1+{{\tan }^{2}}x)}^{2}}(\operatorname{t})dt} \\\ & \int{{{(1+{{\operatorname{t}}^{2}})}^{2}}(\operatorname{t})dt} \\\ \end{aligned}$
(Since, t = tanx)
Further, expanding the square term, we get,
$\begin{aligned} & \int{(1+{{t}^{4}}+2{{t}^{2}})(\operatorname{t})dt} \\\ & \int{(t+{{t}^{5}}+2{{t}^{3}})dt}--(B) \\\ \end{aligned}$
Now, we should know the basics of integration of x raised to the power n. Thus, we should be aware of-
$\int{{{t}^{n}}dt}=\dfrac{{{t}^{n+1}}}{n+1}$
Where, n is any integer with n$\ne$1.
We use this formula in (B), we get,
$\begin{aligned} & \int{(t+{{t}^{5}}+2{{t}^{3}})dt} \\\ & \left[ \dfrac{{{t}^{2}}}{2}+\dfrac{{{t}^{6}}}{6}+\dfrac{2{{t}^{4}}}{4} \right]+c \\\ \end{aligned}$
We have an additional term c; this is an integration constant which we get while performing indefinite integration. This term arrives because while doing the differentiation, C term vanishes since the differentiation of integration constant is zero.
Now we substitute t=tanx (which we earlier took). We get,
$\left[ \dfrac{{{\tan }^{2}}x}{2}+\dfrac{{{\tan }^{6}}x}{6}+\dfrac{2{{\tan }^{4}}x}{4} \right]+c$ -- (C)

Note: Another way to solve the problem is substituting t = sec(x). Then, we have dt = sec(x)tan(x)dx. Thus, we have,
$\int{{{\sec }^{6}}x\tan xdx}$
Now, by re-arranging, we have,
$\int{{{\sec }^{5}}x(\sec x\tan x)dx}$
Now, we have dt = sec(x)tan(x)dx and t = sec(x). Thus,
$\int{{{t}^{5}}dt}$ (Since, ${{\sec }^{5}}x={{t}^{5}}$)
$\dfrac{{{t}^{6}}}{6}$ = $\dfrac{{{\sec }^{6}}x}{6}$ (Since, t = sec(x) )
Hence, the correct answer is $\dfrac{{{\sec }^{6}}x}{6}+c$.