Question

How do you integrate $\int{{{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx}$ ?

Answer 1

We are asked to find the integral of ${{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx$ .To do so we learn how integral work, how are the various functions are connected, how to integrate the basic functions.
Then we see that how terms of our function ${{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)$ are connected.
We will use that $\tan x$ derivative is ${{\sec }^{2}}x$ , so we use this to solve our problem,
We will substitute ‘t’ as $\tan \left( x \right)$ then we find value of ‘dx’ and solve further, lastly we use integral formula $\int{{{x}^{n}}dx=\left( \dfrac{{{x}^{x+1}}}{n+1} \right)}$ to compute the integral.

Complete step by step answer:
We are given ${{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)$ . We are asked to integrate it.
Before we start solving our problem, we must learn how the integrator works.
We know that integration is just the opposite of differentiation.
In differentiation we separate things to smaller pieces while in integrators we add small things which lead us to bigger pieces.
Basic integration formula are $\int{{{x}^{n}}dx=\left( \dfrac{{{x}^{n+1}}}{n+1} \right)}$
Power is being increased by one and simultaneously the result is divided by $n+1$ .
We are asked to integrate $\int{{{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx}$.
Here we can see that $\sec \left( x \right)$ and $\tan \left( x \right)$ are connected to each other.
We know that derivative of $\tan x$ is ${{\sec }^{2}}\left( x \right)$ so we use this properties of $\sec \left( x \right)$ and $\tan \left( x \right)$ to solve problems.
We have to integrate ${{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx$ .
We start by substituting $\tan x$ as ‘t’
So $\tan x=t$
By differentiating both sides, we get –
$\dfrac{d\left( \tan \left( x \right) \right)}{dx}=\dfrac{dt}{dx}$
So using this we get –
$dx=\dfrac{dt}{{{\sec }^{2}}x}$ …………………. (1)
Now we use this in $\int{{{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx}$
Now we use $dx=\dfrac{dt}{{{\sec }^{2}}\left( x \right)}$ and $\tan x=t$
So, $\int{{{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx}=\int{{{\sec }^{2}}\left( x \right){{t}^{3}}\times \dfrac{dt}{{{\sec }^{2}}\left( x \right)}}$
Now simplifying, we get –
$=\int{{{t}^{3}}dt}$
Now we apply $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ to integrate further.
So, $=\int{{{t}^{3}}dt=\left( \dfrac{{{t}^{3+1}}}{3+1} \right)}$
By simplifying, we get –
$=\dfrac{{{t}^{4}}}{4}+c$
Now we substitute back the value of ‘t’ as $\tan x$ so, we get –
$=\dfrac{{{t}^{4}}}{4}+c=\dfrac{{{\tan }^{4}}x}{4}+c$ where ‘c’ is constant.
So, finally we get –
$\int{{{\sec }^{2}}\left( x \right){{\tan }^{3}}\left( x \right)dx}=\dfrac{{{\tan }^{4}}\left( x \right)}{4}+c$

Note: The integration of constant is also find using the identity $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ , if we have $\int{1dx}$ then 1 can be written as ‘x’ so $\int{1dx=\int{{{x}^{0}}dx}}$ which give us $\dfrac{{{x}^{0+1}}}{0+1}$ .
By simplifying, we get ‘x’.
Always if we have an indefinite integral (integral without limit) then we have to change the substitution term back to the original after the complete solution, like we change ‘t’ back as tan x. But if we have definite integral (integral with limit) then, we can change limits according to the substitution and we do not need to come back.