Question

How do you evaluate the definite integral $\int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx$ from $1$ to $\sqrt 5$?

Answer 1

Hint : We have to find the definite integral of $\int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx$ from $1$ to $\sqrt 5$ i.e., $\int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx$. For this we will use the substitution method. We will substitute ${x^2} - 1$ by $u$ and then by using the basic integration formula we will integrate it to find the result. In the final result substitute the original variable back before putting the limits else solution will be wrong.

Complete step-by-step answer :
Definite integrals are those integrals which have limit of integration i.e., upper limit and lower limit. It has two different values for upper limit and lower limit when they are evaluated.
We have to evaluate the definite integral $\int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx$ from $1$ to $\sqrt 5$. We can also write it as $\int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx$.
Let $I = \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx - - - (1)$.
We can make this integration easier by using some substitution. We will substitute ${x^2} - 1$ by $u$.
Let ${x^2} - 1 = u$
On differentiating with respect to $x$, we get
$\Rightarrow 2x = \dfrac{{du}}{{dx}}$
On rearranging, we get
$\Rightarrow du = 2xdx$
On substitution $(1)$ becomes,
$I = \int {\dfrac{{du}}{{\sqrt u }}}$
On rewriting we get
$I = \int {{u^{\dfrac{{ - 1}}{2}}}du}$
On integration we get
$I = \dfrac{{{u^{\dfrac{{ - 1}}{2} + 1}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}}$
On simplification we get
$I = 2\sqrt u$
Putting $u = {x^2} - 1$, we get
$I = 2\left( {\sqrt {{x^2} - 1} } \right)$
Putting the upper of $\sqrt 5$ and lower limit of $1$, we get
$\Rightarrow \int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 2\left[ {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - 1} - \sqrt {{{\left( {\sqrt 1 } \right)}^2} - 1} } \right]$
On simplification we get
$\Rightarrow \int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 2\left[ {\left( {\sqrt {5 - 1} } \right) - \left( {\sqrt {1 - 1} } \right)} \right]$
On further simplification we get
$\Rightarrow \int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 2\left[ {\left( {\sqrt {{{\left( 2 \right)}^2}} } \right) - 0} \right]$
On simplifying the term in the bracket, we get
$\Rightarrow \int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 2 \times 2$
On simplification, we get
$\Rightarrow \int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 4$
Therefore, the value of the given definite integral $\int\limits_1^{\sqrt 5 } {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}} dx = 4$.

Note : In definite integrals, upper and lower limits are given so after integration we just get a number whereas in indefinite integrals, we just integrate a function and add an arbitrary constant. The final value of a definite integral is the value of integral for the upper limit minus value of the integral for the lower limit.