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

Hint:
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.