Question

Find the value of the following integral.
\[\int\limits_0^1 {4{x^3}} \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx\]

Answer 1

Hint: In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process that is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$ , where C is the integration constant.
Use integration by parts in question involving the integration of the product of two functions.
Integration by parts
        “ The integral of the product of two functions = (first function) x (integral of the second function) – Integral of [differential coefficient of the first function ) x (integral of the second function)]”
i.e. \[\int {{f_1}\left( x \right)} \cdot {f_2}\left( x \right) = {f_1}\left( x \right)\int {{f_2}\left( x \right)} - \int {\left( {{f_1}^\prime \left( x \right) \cdot {f_2}\left( x \right)} \right)} dx\]
Tip: by ILATE choose the first function (Preference :: Inverse Trigonometric > Logarithmic Function > Algebra > Trigonometric Functions > Exponential )
The definite integral is calculated whenever a function \[f\] is given over the range $\left[ {a,b} \right]$ , where $a$ and $b$ are called limits of integration, $a$ being the lower limit and $b$ being the upper limit. I.e.:
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$
In other words, definite integral (integral whose limits are given) is given by upper limits minus lower limits.

Complete step-by-step answer:
Let given definite Integral
\[I = \int\limits_0^1 {4{x^3}} \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx\]
Using integration by parts: \[\int {{f_1}\left( x \right)} \cdot {f_2}\left( x \right) = {f_1}\left( x \right)\int {{f_2}\left( x \right)} - \int {\left( {{f_1}^\prime \left( x \right) \cdot {f_2}\left( x \right)} \right)} dx\]
Take \[{f_1}\left( x \right) = 4{x^3};{f_2}\left( x \right) = \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}\]
         \[{f_1}^\prime \left( x \right) = 4 \times 3{x^2} = 12{x^2}\] (using differentiation: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}$ )
        $\int {{f_2}\left( x \right)} = \int {\dfrac{{{d^2}}}{{d{x^2}}}} {\left( {1 - {x^2}} \right)^5} = \dfrac{d}{{dx}}{\left( {1 - {x^2}} \right)^5}$
Thus, $I = \left. {\left[ {4{x^3}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right) - \int {\left\{ {12{x^2}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right)} \right\}dx} } \right]} \right|_0^1$
Solving differentiation: $\dfrac{d}{{dx}}{\left( {1 - {x^2}} \right)^5}$
Using differential: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}$
  $\dfrac{d}{{dx}}{\left( {1 - {x^2}} \right)^5}$
$ \Rightarrow 5{\left( {1 - {x^2}} \right)^4}\left( { - 2x} \right)$
Hence, $I = \left. {\left[ {4{x^3}\left( {5{{\left( {1 - {x^2}} \right)}^4}\left( { - 2x} \right)} \right) - \int {\left\{ {12{x^2}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right)} \right\}dx} } \right]} \right|_0^1$
               $I = \left. {\left[ { - 40{x^4}{{\left( {1 - {x^2}} \right)}^4} - \int {\left\{ {12{x^2}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right)} \right\}dx} } \right]} \right|_0^1$
Solve integral $\int {\left\{ {12{x^2}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right)} \right\}dx} $ separately to reduce errors in calculation.
Let ${I_1} = \int {\left\{ {12{x^2}\left( {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right)} \right\}dx} $
Using integration by parts: \[\int {{f_1}\left( x \right)} \cdot {f_2}\left( x \right) = {f_1}\left( x \right)\int {{f_2}\left( x \right)} - \int {\left( {{f_1}^\prime \left( x \right) \cdot {f_2}\left( x \right)} \right)} dx\]
Take \[{f_1}\left( x \right) = 12{x^2};{f_2}\left( x \right) = \left\{ {\dfrac{d}{{dx}}{{\left( {1 - {x^2}} \right)}^5}} \right\}\]
        \[{f_1}^\prime \left( x \right) = 12 \times 2x = 24x\] (using differentiation: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}$ )
       $\int {{f_2}\left( x \right)} = \int {\dfrac{d}{{dx}}} {\left( {1 - {x^2}} \right)^5} = {\left( {1 - {x^2}} \right)^5}$
Thus, ${I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \int {24x{{\left( {1 - {x^2}} \right)}^5}} dx$
Using binomial expansion expand: ${\left( {1 - {x^2}} \right)^5}$
${\left( {a + b} \right)^n} = C\left( {n,0} \right){a^n} + C\left( {n,1} \right){a^{n - 1}}b + C\left( {n,2} \right){a^{n - 2}}{b^2} + ... + C(n,n){b^n}$
On comparing ${\left( {1 - {x^2}} \right)^5}$with ${\left( {a + b} \right)^n}$
$
  a = 1; \\
  b = \left( { - {x^2}} \right); \\
  n = 5 \\
 $
$
  {\left( {1 + \left( { - {x^2}} \right)} \right)^5} \\
   = C\left( {5,0} \right){1^5} + C\left( {5,1} \right){1^4}\left( { - {x^2}} \right) + C\left( {5,2} \right){1^3}{\left( { - {x^2}} \right)^2} + C\left( {5,3} \right){1^2}{\left( { - {x^2}} \right)^3} + C\left( {5,4} \right){1^1}{\left( { - {x^2}} \right)^4} + C\left( {5,5} \right){\left( { - {x^2}} \right)^5} \\
 $
We know, combination: $C(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$C\left( {5,0} \right) = C\left( {5,5} \right) = 1$
$C\left( {5,1} \right) = C\left( {5,4} \right) = 5$
$C\left( {5,2} \right) = C\left( {5,3} \right) = 10$
Hence, ${\left( {1 + \left( { - {x^2}} \right)} \right)^5} = 1 - 5{x^2} + 10{x^4} - 10{x^6} + 5{x^8} - {x^{10}}$
Thus, substituting in integral ${I_1}$
${I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \int {24x\left( {1 - 5{x^2} + 10{x^4} - 10{x^6} + 5{x^8} - {x^{10}}} \right)} dx$
\[{I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \int {\left( {24x - 120{x^3} + 240{x^5} - 240{x^7} + 120{x^9} - 24{x^{11}}} \right)} dx\]
Using the integration: $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
\[{I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \left( {24\dfrac{{{x^2}}}{2} - 120\dfrac{{{x^4}}}{4} + 240\dfrac{{{x^6}}}{6} - 240\dfrac{{{x^8}}}{8} + 120\dfrac{{{x^{10}}}}{{10}} - 24\dfrac{{{x^{12}}}}{{12}}} \right)\]
\[{I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \left( {12{x^2} - 30{x^4} + 40{x^6} - 30{x^8} + 12{x^{10}} - 2{x^{12}}} \right)\]
Now substitute ${I_1}$
We know, $I = \left. {\left[ { - 40{x^4}{{\left( {1 - {x^2}} \right)}^4} - {I_1}} \right]} \right|_0^1$

Substituting the integral ${I_1}$
\[I = \left. {\left[ { - 40{x^4}{{\left( {1 - {x^2}} \right)}^4} - \left\{ {12{x^2}{{\left( {1 - {x^2}} \right)}^5} - \left( {12{x^2} - 30{x^4} + 40{x^6} - 30{x^8} + 12{x^{10}} - 2{x^{12}}} \right)} \right\}} \right]} \right|_0^1\]
\[I = \left. {\left[ { - 40{x^4}{{\left( {1 - {x^2}} \right)}^4} - 12{x^2}{{\left( {1 - {x^2}} \right)}^5} + 12{x^2} - 30{x^4} + 40{x^6} - 30{x^8} + 12{x^{10}} - 2{x^{12}}} \right]} \right|_0^1\]
Solving limits of the integral:
\[
  I = \left[ { - 40{{\left( 1 \right)}^4}{{\left( {1 - {1^2}} \right)}^4} - 12{{\left( 1 \right)}^2}{{\left( {1 - {1^2}} \right)}^5} + 12{{\left( 1 \right)}^2} - 30{{\left( 1 \right)}^4} + 40{{\left( 1 \right)}^6} - 30{{\left( 1 \right)}^8} + 12{{\left( 1 \right)}^{10}} - 2{{\left( 1 \right)}^{12}}} \right] - \\
  {\text{ }}\left[ { - 40{{\left( 0 \right)}^4}{{\left( {1 - {0^2}} \right)}^4} - 12{{\left( 0 \right)}^2}{{\left( {1 - {0^2}} \right)}^5} + 12{{\left( 0 \right)}^2} - 30{{\left( 0 \right)}^4} + 40{{\left( 0 \right)}^6} - 30{{\left( 0 \right)}^8} + 12{{\left( 0 \right)}^{10}} - 2{{\left( 0 \right)}^{12}}} \right] \\
 \]
$
   \Rightarrow 0 - 0 + 12 - 30 + 40 - 30 + 12 - 2 \\
   \Rightarrow 64 - 62 \\
   \Rightarrow 2 \\
 $
The value of \[\int\limits_0^1 {4{x^3}} \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx\] is 2.

Note:
Carefully do the calculation, emphasize on each and every step.
Steps involving many terms and variables of different power should be dealt with carefully.
Students are likely to make mistakes while substituting ${I_1}$in $I = \left. {\left[ { - 40{x^4}{{\left( {1 - {x^2}} \right)}^4} - {I_1}} \right]} \right|_0^1$. Notice that there is a minus sign before ${I_1}$. The calculation will go wrong if you ignore it.
In the step: \[{I_1} = 12{x^2}{\left( {1 - {x^2}} \right)^5} - \int {\left( {24x - 120{x^3} + 240{x^5} - 240{x^7} + 120{x^9} - 24{x^{11}}} \right)} dx\], we have not expanded \[{\left( {1 - {x^2}} \right)^5}\] as it is not included in integral, we can leave it as it is and find limits when requires.

Alternate method:
\[I = \int\limits_0^1 {4{x^3}} \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx\]
The part \[\dfrac{{{d^2}}}{{d{x^2}}}{\left( {1 - {x^2}} \right)^5}\] in given integral is the double differentiation of the function \[{\left( {1 - {x^2}} \right)^5}\] .
Therefore, solve the differentiation before and substitute it in the given integral.
Let \[y = {\left( {1 - {x^2}} \right)^5}\]
On differentiating both sides with respect to $x$ .
Using differential: $\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{dy}}{{dx}} = 5{\left( {1 - {x^2}} \right)^4}\left( { - 2x} \right)$
$\dfrac{{dy}}{{dx}} = - 10x{\left( {1 - {x^2}} \right)^4}$
On differentiating both sides with respect to $x$ .
Using product rule: $\dfrac{d}{{dx}}u\left( x \right) \cdot v\left( x \right) = u'\left( x \right) \cdot v\left( x \right) + u\left( x \right) \cdot v'\left( x \right)$
$\dfrac{{{d^2}y}}{{d{x^2}}} = - 10{\left( {1 - {x^2}} \right)^4} - 10x\left( 4 \right){\left( {1 - {x^2}} \right)^3}\left( { - 2x} \right)$
$ \Rightarrow - 10{\left( {1 - {x^2}} \right)^4} + 80{x^2}{\left( {1 - {x^2}} \right)^3}$
Expanding ${\left( {1 - {x^2}} \right)^3}$ using identity: ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2}$
On comparing ${\left( {1 - {x^2}} \right)^3}$with ${\left( {a - b} \right)^3}$
$
  a = 1; \\
  b = {x^2} \\
 $
${\left( {1 - {x^2}} \right)^3} = {1^3} - {\left( {{x^2}} \right)^3} - 3{\left( 1 \right)^2}{x^2} + 3\left( 1 \right){\left( {{x^2}} \right)^2}$
$ \Rightarrow 1 - {x^6} - 3{x^2} + 3{x^4}$
Expanding ${\left( {1 - {x^2}} \right)^4}$using binomial expansion:
${\left( {a + b} \right)^n} = C\left( {n,0} \right){a^n} + C\left( {n,1} \right){a^{n - 1}}b + C\left( {n,2} \right){a^{n - 2}}{b^2} + ... + C(n,n){b^n}$
On comparing ${\left( {1 - {x^2}} \right)^4}$with ${\left( {a + b} \right)^n}$
$
  a = 1; \\
  b = \left( { - {x^2}} \right); \\
  n = 4 \\
 $
$
  {\left( {1 + \left( { - {x^2}} \right)} \right)^4} \\
   = C\left( {4,0} \right){1^4} + C\left( {4,1} \right){1^3}\left( { - {x^2}} \right) + C\left( {4,2} \right){1^2}{\left( { - {x^2}} \right)^2} + C\left( {4,3} \right){1^1}{\left( { - {x^2}} \right)^3} + C\left( {4,4} \right){\left( { - {x^2}} \right)^4} \\
 $
We know, combination: $C(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$C\left( {4,0} \right) = C\left( {4,4} \right) = 1$
$C\left( {4,1} \right) = C\left( {4,3} \right) = 4$
$C\left( {4,2} \right) = 6$
${\left( {1 + \left( { - {x^2}} \right)} \right)^4} = 1 - 4{x^2} + 6{x^4} - 4{x^6} + {x^8}$
On substituting expansion of ${\left( {1 - {x^2}} \right)^4}$ and ${\left( {1 - {x^2}} \right)^3}$ in $\dfrac{{{d^2}y}}{{d{x^2}}}$
$\dfrac{{{d^2}y}}{{d{x^2}}} = - 10{\left( {1 - {x^2}} \right)^4} + 80{x^2}{\left( {1 - {x^2}} \right)^3}$
$
   \Rightarrow - 10\left( {1 - 4{x^2} + 6{x^4} - 4{x^6} + {x^8}} \right) + 80{x^2}\left( {1 - {x^6} - 3{x^2} + 3{x^4}} \right) \\
   \Rightarrow - 10 + 40{x^2} - 60{x^4} + 40{x^6} - 10{x^8} + 80{x^2} - 80{x^8} - 240{x^4} + 240{x^6} \\
   \Rightarrow - 10 + 120{x^2} - 300{x^4} + 280{x^6} - 90{x^8} \\
 $
Substituting $\dfrac{{{d^2}y}}{{d{x^2}}}$ in given integral:

\[I = \int\limits_0^1 {4{x^3}} \left\{ {\dfrac{{{d^2}}}{{d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx\]
$I = \int\limits_0^1 {\left[ {4{x^3}\left( { - 10 + 120{x^2} - 300{x^4} + 280{x^6} - 90{x^8}} \right)} \right]} dx$
$ \Rightarrow \int\limits_0^1 {\left[ { - 40{x^3} + 480{x^5} - 1200{x^7} + 1120{x^9} - 360{x^{11}}} \right]} dx$
Using the integration: $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
\[
  I = \left. {\left[ { - 40\dfrac{{{x^4}}}{4} + 480\dfrac{{{x^6}}}{6} - 1200\dfrac{{{x^8}}}{8} + 1120\dfrac{{{x^{10}}}}{{10}} - 360\dfrac{{{x^{12}}}}{{12}}} \right]} \right|_0^1 \\
   \Rightarrow \left. {\left[ { - 10{x^4} + 80{x^6} - 150{x^8} + 112{x^{10}} - 30{x^{12}}} \right]} \right|_0^1 \\
 \]
On solving limits:
\[
  I = \left[ { - 10{{\left( 1 \right)}^4} + 80{{\left( 1 \right)}^6} - 150{{\left( 1 \right)}^8} + 112{{\left( 1 \right)}^{10}} - 30{{\left( 1 \right)}^{12}}} \right] - \\
  {\text{ }}\left[ { - 10{{\left( 0 \right)}^4} + 80{{\left( 0 \right)}^6} - 150{{\left( 0 \right)}^8} + 112{{\left( 0 \right)}^{10}} - 30{{\left( 0 \right)}^{12}}} \right] \\
 \]
$
  I = - 10 + 80 - 150 + 112 - 30 \\
   \Rightarrow 192 - 190 \\
   \Rightarrow 2 \\
 $