Question

Evaluate the integral $\int {\sqrt {{x^2} - 1} } dx$.

Answer 1

To solve this problem, we must know how to integrate by parts and the formula for this is $\int {udv} = uv - \int {vdu}$. You must be clear in the concept indefinite integral because the limits were not given in this problem. So, you must know how to deal with indefinite integral. Some of the other formula you need to know is and
$\int {\dfrac{1}{{\sqrt {{x^2} - 1} }}} dx$ $=$ $\log \left( {x + \sqrt {{x^2} - 1} } \right) + c$

Complete step by step answer:
Let us consider the given problem,
$\int {\sqrt {{x^2} - 1} } dx$
Take $I = \sqrt {{x^2} - 1} dx$, integrate this by parts and formula for this is $\int {udv} = uv - \int {vdu}$, let’s take $u = \sqrt {{x^2} - 1}$ and to find $du$ we need to differentiate $u$,
$du = \dfrac{1}{{2\sqrt {{x^2} - 1} }}\left( {2x - 0} \right)dx = \dfrac{x}{{\sqrt {{x^2} - 1} }}$
Let’s take $dv = dx$, if we integrate this we get $v$,
$v = x$

Substituting the value of $u$, $du$, $v$ and $dv$ we get,
$\Rightarrow I = x\sqrt {{x^2} - 1} - \int x \left( {\dfrac{x}{{\sqrt {{x^2} - 1} }}} \right)dx \\\ \Rightarrow I = x\sqrt {{x^2} - 1} - \int {\dfrac{{{x^2}}}{{\sqrt {{x^2} - 1} }}} dx \\\$
Add and subtract with $1$ in the second part of the above equation we get,
$\Rightarrow I = x\sqrt {{x^2} - 1} - \int {\dfrac{{{x^2} - 1 + 1}}{{\sqrt {{x^2} - 1} }}} dx$
We are writing the above equation as,
$\Rightarrow I = x\sqrt {{x^2} - 1} - \int {\left( {\dfrac{{{x^2} - 1}}{{\sqrt {{x^2} - 1} }} + \dfrac{1}{{\sqrt {{x^2} - 1} }}} \right)} dx$
$\Rightarrow \int {\dfrac{{{x^2} - 1}}{{\sqrt {{x^2} - 1} }}}$ becomes $\int {\dfrac{{{x^2} - 1}}{{\sqrt {{x^2} - 1} }} = \int {\dfrac{{\sqrt {{x^2} - 1} \times \sqrt {{x^2} - 1} }}{{\sqrt {{x^2} - 1} }}} } = \int {\sqrt {{x^2} - 1} }$, the above equation becomes,
$\Rightarrow I = x\sqrt {{x^2} - 1} - \int {\sqrt {{x^2} - 1} } dx - \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}} dx$
We know that, $I = \sqrt {{x^2} - 1} dx$ and $\int {\dfrac{1}{{\sqrt {{x^2} - 1} }}} dx$ $=$ $\log \left( {x + \sqrt {{x^2} - 1} } \right) + c$
$\Rightarrow I = x\sqrt {{x^2} - 1} - I - \log \left( {x + \sqrt {{x^2} - 1} } \right) + c$
$\Rightarrow I + I = x\sqrt {{x^2} - 1} - \log \left( {x + \sqrt {{x^2} - 1} } \right) + c$
$\Rightarrow 2I = x\sqrt {{x^2} - 1} - \log \left( {x + \sqrt {{x^2} - 1} } \right) + c$
$\Rightarrow I = \dfrac{x}{2}\sqrt {{x^2} - 1} - \dfrac{1}{2}\log \left( {x + \sqrt {{x^2} - 1} } \right) + \dfrac{c}{2}$
This is our required solution.

Additional information: This problem is one of the difficult problems in indefinite integral because only when you know the formula for the integral, which I have mentioned in the note, you are able to solve this problem. And integrating by parts method is a common method which is used most common in major problems in integration.

Note: To prove $\int {\dfrac{1}{{\sqrt {{x^2} - 1} }}} dx$ $=$ $\log \left( {x + \sqrt {{x^2} - 1} } \right) + c$, take $u = x + \sqrt {{x^2} - {1^2}}$ and by differentiating this we get $\Rightarrow du = \left( {1 + \dfrac{{2x}}{{2\sqrt {{x^2} - {1^2}} }}} \right)dx$
$\Rightarrow du = \left( {\dfrac{{\sqrt {{x^2} - {1^2}} + x}}{{\sqrt {{x^2} - {1^2}} }}} \right)dx$
$\Rightarrow dx = \left( {\dfrac{{\sqrt {{x^2} - {1^2}} }}{{\sqrt {{x^2} - {1^2}} + x}}} \right)du$
Substituting the value of $dx$ and $u$ in $I$ we get,
$\Rightarrow I = \int {\left( {\dfrac{1}{{\sqrt {{x^2} - 1} }} \times \dfrac{{\sqrt {{x^2} - {1^2}} }}{u}} \right)} du$
$\Rightarrow I = \int {\dfrac{{du}}{u}}$
$I = \log u + c$, we already know that $u = x + \sqrt {{x^2} - {1^2}}$ so in the question we directly wrote this integral. Know this value so that you are able to save time in the examination.