Question

Write the integral of $\sqrt {{x^2} - {a^2}}$ with respect to $x$, and hence evaluate $\int {\sqrt {{x^2} - 8x + 7} } dx$.

Answer 1

Here, we have to evaluate the given integral. We will use completing the square method to express the given integral in the form $\sqrt {{x^2} - {a^2}}$. Then, using the formula for integral of $\sqrt {{x^2} - {a^2}}$, we will evaluate the value of the required integral.

Formula Used: We will use the following formulas:

1. The value of the integral of $\sqrt {{x^2} - {a^2}}$ with respect to $x$ is given as $\int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$, where $C$ is a constant of integration.
2. The square of the difference of two numbers is given by the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.

Complete step by step solution:
First, we need to write the integral of $\sqrt {{x^2} - {a^2}}$ with respect to $x$.
Now, we need to use this formula to evaluate the integral $\int {\sqrt {{x^2} - 8x + 7} } dx$.
We will use completing the square method to express the integral $\int {\sqrt {{x^2} - 8x + 7} } dx$ in the form $\sqrt {{x^2} - {a^2}}$.
To complete the square, we need to add and subtract the half of the coefficient of $x$.
The coefficient of $x$ is 8. The half of 8 is 4.
We will add and subtract the square of 4 in the expression under the root.
Therefore, the integral becomes
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{x^2} - 8x + 7 + {{\left( 4 \right)}^2} - {{\left( 4 \right)}^2}} } dx$
Rewriting the expression, we get
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{{\left( x \right)}^2} - 2\left( x \right)\left( 4 \right) + {{\left( 4 \right)}^2} + 7 - {{\left( 4 \right)}^2}} } dx$
We will use the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ to simplify the expression.
Substituting $a = x$ and $b = 4$ in the identity, we get
$\Rightarrow {\left( {x - 4} \right)^2} = {x^2} - 2\left( x \right)\left( 4 \right) + {\left( 4 \right)^2}$
Substituting ${x^2} - 2\left( x \right)\left( 4 \right) + {\left( 4 \right)^2} = {\left( {x - 4} \right)^2}$ in the value of the integral, we get
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{{\left( {x - 4} \right)}^2} + 7 - {{\left( 4 \right)}^2}} } dx$
Applying the exponent on the base, we get
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{{\left( {x - 4} \right)}^2} + 7 - 16} } dx$
Subtracting 16 from 7, we get
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{{\left( {x - 4} \right)}^2} - 9} } dx$
Rewriting 9 as the square of 3, we get
$\Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \int {\sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} } dx$
We can observe that this integral is of the form $\sqrt {{x^2} - {a^2}}$.
The value of the integral of $\sqrt {{x^2} - {a^2}}$ with respect to $x$ is given as $\int {\sqrt {{x^2} - {a^2}} } dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}} - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$, where $C$ is a constant of integration.
Substituting $x - 4$ for $x$, and 3 for $a$ in the formula, we get
$\Rightarrow \int {\sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} } dx = \dfrac{{\left( {x - 4} \right)}}{2}\sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} - \dfrac{{{3^2}}}{2}\log \left| {\left( {x - 4} \right) + \sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} } \right| + C$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \dfrac{{x - 4}}{2}\sqrt {{x^2} - 8x + 7} - \dfrac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\\\ \Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \dfrac{x}{2}\sqrt {{x^2} - 8x + 7} - \dfrac{4}{2}\sqrt {{x^2} - 8x + 7} - \dfrac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\\\ \Rightarrow \int {\sqrt {{x^2} - 8x + 7} } dx = \dfrac{x}{2}\sqrt {{x^2} - 8x + 7} - 2\sqrt {{x^2} - 8x + 7} - \dfrac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\end{array}$
Therefore, the value of the integral $\int {\sqrt {{x^2} - 8x + 7} } dx$ is $\dfrac{x}{2}\sqrt {{x^2} - 8x + 7} - 2\sqrt {{x^2} - 8x + 7} - \dfrac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C$.

Note:
To complete the square, we added and subtracted the half of the coefficient of $x$. This is to be done only if the coefficient of ${x^2}$ is 1. If the coefficient of ${x^2}$ is not 1, divide the expression such that the coefficient of ${x^2}$ is 1. Then, add and subtract the half of the coefficient of $x$. The number used to divide is a constant and can be taken outside the integral.