Question

Evaluate the following integral: \[\int {2x{{\sec }^3}({x^2} + 3)\tan ({x^2} + 3)dx} \]?

Answer 1

Hint: We can try to solve this question by using the substitution method. We can assign a variable, a part of the question, and then start differentiating it. This makes the question complex-free and it is easy to solve afterwards. Then we can try to put that differentiated part in the question back again and integrate it.

Complete step-by-step solution:
The integral is:
\[\int {2x{{\sec }^3}({x^2} + 3)\tan ({x^2} + 3)dx} \]
We are going to integrate this integral with respect to \[x\].
First, we will try to expand \[{\sec ^3}\] and \[\tan \]will remain the same. We can expand it in \[{\sec ^2}\,\,and\,\,\sec \]and then, we will get:
\[ = \int {2x{{\sec }^2}({x^2} + 3)\sec ({x^2} + 3)\tan ({x^2} + 3)dx} \]
Now, we will assign a variable to \[\sec ({x^2} + 3)\]. So, let \[t = \sec ({x^2} + 3)\]. Now, we will differentiate \[t\], and we will get:
\[\Rightarrow \dfrac{d}{{dx}}t = \dfrac{d}{{dx}}\sec ({x^2} + 3)\]
We know that the derivation of \[\sec x\]is \[\sec x\tan x\]:
\[\Rightarrow \dfrac{d}{{dx}}(\sec x) = \sec x\tan x\]
If the derivation of \[\sec x\]is \[\sec x\tan x\], then the derivation of \[\sec ({x^2} + 3)\]is:
\[\Rightarrow \dfrac{d}{{dx}}\sec ({x^2} + 3) = \sec ({x^2} + 3)\tan ({x^2} + 3) \cdot \dfrac{d}{{dx}}({x^2} + 3)\]
We know from the basic rule of differentiation that:
\[\Rightarrow \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
We will apply this rule to differentiate \[\dfrac{d}{{dx}}({x^2} + 3)\], and we get:
\[\Rightarrow \dfrac{d}{{dx}}({x^2} + 3) = \dfrac{d}{{dx}}({x^2}) + \dfrac{d}{{dx}}(3)\]
\[ \Rightarrow \dfrac{d}{{dx}}({x^2} + 3) = 2x\]
When we put the value of \[\dfrac{d}{{dx}}({x^2} + 3)\]in the equation \[\dfrac{d}{{dx}}\sec ({x^2} + 3) = \sec ({x^2} + 3)\tan ({x^2} + 3) \cdot \dfrac{d}{{dx}}({x^2} + 3)\], then we get:
\[ \Rightarrow \dfrac{d}{{dx}}\sec ({x^2} + 3) = \sec ({x^2} + 3)\tan ({x^2} + 3) \cdot 2x\]
So, differentiation of \[t = \sec ({x^2} + 3)\]is:
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \sec ({x^2} + 3)\tan ({x^2} + 3) \cdot 2x\]
Now, we will shift the \[dx\]to the other side of the equation. This is done so that we can integrate the equation. After shifting \[dx\], we get:
\[ \Rightarrow dt = \left[ {\sec ({x^2} + 3)\tan ({x^2} + 3) \cdot 2x} \right]dx\]
When we rearrange the equation, we get:
\[ \Rightarrow dt = \left[ {2x\sec ({x^2} + 3)\tan ({x^2} + 3)} \right]dx\]
Now, when we replace the value of \[dt\] in our question, we get:
\[ \Rightarrow \int {{{\sec }^2}({x^2} + 3)dt} \]
We have assigned earlier that \[t = \sec ({x^2} + 3)\]. Now, if we replace the value of \[t\]in the question, then we get:
\[ \Rightarrow \int {{t^2}dt} \]
This has made our integration process a lot easier. Now, we will apply the basic rule of integration that is:
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \]
When we solve \[\int {{t^2}dt} \]according to the formula, then we get:
 \[ \Rightarrow \int {{t^2}dt} = \dfrac{{{t^{2 + 1}}}}{{2 + 1}}\]
\[ \Rightarrow \int {{t^2}dt} = \dfrac{{{t^3}}}{3} + C\]
Therefore, we got \[\dfrac{{{t^3}}}{3} + C\]
Now, we will simply put the value of \[t = \sec ({x^2} + 3)\]in the equation, and we will get our answer as:
\[ = \dfrac{{{{\sec }^3}({x^2} + 3)}}{3} + C\]

Therefore, \[\int {2x{{\sec }^3}({x^2} + 3)\tan ({x^2} + 3)dx} = \dfrac{{{{\sec }^3}({x^2} + 3)}}{3} + C\]

Note: In this type of question, we should always use the substitution method. By substituting, it becomes a lot easier to differentiate and integrate in the later part of the solution. But, if the expression is big, then it is more confusing, so we have to substitute it correctly.