Question

How do you integrate \[\int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}} \] from \[ - 3\] to \[ - 2\] ?

Answer 1

Hint: At first, we will solve for the indefinite integration as \[\int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}} \] .
Then we will put the limit of \[x.\]
To solve the indefinite integration, we will use a substitute method.
After putting the limit, we will get the final answer.

Complete step by step answer:
We have to find the value of \[\int\limits_{ - 3}^{ - 2} {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}} \]
At first, we will solve for the indefinite integration.
\[ \Rightarrow \int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}} \]
We will solve it by substitution method.
Let us take, \[x = \sec \theta \]
Differentiate both sides with respect to \[x\] we get,
\[dx = \sec \theta \tan \theta d\theta \]
Substituting these values in the given integration we get,
\[ \Rightarrow \int {\dfrac{{\sec \theta \tan \theta d\theta }}{{\sqrt {{{\sec }^2}\theta - 1} }}} \]
Simplifying we get,
\[ \Rightarrow \int {\sec \theta d\theta } \]
Integrating we get,
\[ \Rightarrow \log \left| {\tan \theta + \sec \theta } \right|\]
Substituting the value of \[\theta \] we get,
\[ \Rightarrow \log \left| {\tan (\operatorname{arc} \sec x) + \sec (\operatorname{arc} \sec x)} \right|\]
Simplifying we get,
\[ \Rightarrow \log \left| {\sqrt {{x^2} - 1} + x} \right|\]
Now we will put the limit.
Putting \[x = - 2\] and \[x = - 3\] we get,
\[ \Rightarrow \log \left| {\sqrt {{{( - 2)}^2} - 1} - 2} \right| - \log \left| {\sqrt {{{( - 3)}^2} - 1} - 3} \right|\]
Simplifying we get,
\[ \Rightarrow \log \left| {\sqrt 3 - 2} \right| - \log \left| {\sqrt 8 - 3} \right|\]
On rewriting we get,
\[ \Rightarrow \log \dfrac{{\left| {\sqrt 3 - 2} \right|}}{{\left| {\sqrt 8 - 3} \right|}}\]

Hence, the integration of \[\int {\dfrac{{dx}}{{\sqrt {{x^2} - 1} }}} \] from \[ - 3\] to \[ - 2\] is \[\log \dfrac{{\left| {\sqrt 3 - 2} \right|}}{{\left| {\sqrt 8 - 3} \right|}}\] .

Note: The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly.
In a broad sense, in calculus, the idea of limit is used where algebra and geometry are implemented. Limits help us in the study of the result of points on a graph such as how they get closer to each other until their distance is almost zero.
In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale.
Calculation of small addition problems is an easy task which we can do manually or by using calculators as well. But for big additional problems, where the limits could reach to even infinity, integration methods are used.