# Examples On Integration By Substitution Set-12

The next few examples have integrals of the form we’ve not encountered upto now. But some manipulation can always reduce such integrals to one of the standard forms You are urged to try out these examples on your own before going through the solutions.

**Example - 32 **

Evaluate \(\begin{align}\int {\frac{1}{{{{\sin }^6} + {{\cos }^6}x}}} \,\,dx\end{align}\) .

**Solution: ** One approach could be to convert this expression into one involving tan * x * and sec *x* terms:

\[I = \int {\frac{{{{\sec }^6}x}}{{1 + {{\tan }^6}x}}} \,\,dx\]

We could now use the substitution \(\tan x = t\) so that \({\sec ^2}xdx = dt.\) Thus,

\[\begin{align}& I = \int {\frac{{{{\sec }^4}x\,\,}}{{1 + {t^6}}}dt}\\&\; =\int{\frac{{{\left( 1+{{t}^{2}} \right)}^{2}}\,\,}{1+{{t}^{6}}}dt}\\&\;\; = \int {\frac{{{{\left( {1 + {t^2}} \right)}^2}\,\,}}{{\left( {1 + {t^2}} \right)\left( {1 - {t^2} + {t^4}} \right)}}dt} \,{\text{(By factorizing }}{{\rm{t}}^6}{\text{ + 1 in the denominator)}}\\ &\;\;= \int {\frac{{1 + {t^2}\,\,}}{{1 - {t^2} + {t^4}}}dt}\qquad\qquad\qquad\qquad\qquad\qquad\qquad \,...{\rm{ (1)}}\\ &\;\;= \int {\frac{{1 + \frac{1}{{{t^2}}}\,\,}}{{{t^2} + \frac{1}{{{t^2}}} - 1}}dt} \left\{ \begin{array}{l} {\text{Division of the numerator and the }}\\ {\text{denominator by }}{t^{\rm{2}}}{\rm{,}}\,\,{\text{as in example}}\,\,{\rm{23}} \end{array} \right\} \\& \;\;= \int {\frac{{1 + \frac{1}{{{t^2}}}\,\,}}{{{{\left( {t - \frac{1}{t}} \right)}^2} + 1}}dt}\\&\;\; = \int {\frac{1}{{{y^2} + 1}}dy}\left\{ {{\text{By the substitution }}t - \frac{1}{t} = y} \right\}\\&\;\;= {\tan ^{ - 1}}y + C\\&\;\; = {\tan ^{ - 1}}\left( {t - \frac{1}{t}} \right) + C\\ &\;\; = {\tan ^{ - 1}}\left( {\tan x - \cot x} \right) + C \end{align}\]

An alternative route could be to directly factorize the denominator in the original expression:

\[\begin{align}& {\sin ^6}x + {\cos ^6}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right)\\&\qquad\qquad\qquad = {\sin ^4}x - {\sin ^2}x{\cos ^2}x + {\cos ^4}x\\ &\qquad\qquad\qquad = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x - {\sin ^2}x{\cos ^2}x\\&\qquad\qquad\qquad = 1 - 3{\sin ^2}x{\cos ^2}x\\ &\qquad\quad \Rightarrow\quad I = \int {\frac{1}{{1 - 3{{\sin }^2}x + {{\cos }^2}x}}} \,\,dx\\ & \qquad\qquad\qquad = \int {\frac{{\,{{\sec }^4}x }}{{{{\sec }^4} - 3{{\tan }^2}x}}} \,\,dx \;\;\;\left\{ {{\rm{Division}}\,{\rm{by}}{{\cos }^4}x} \right\} \\ &\qquad\qquad\qquad = \int {\frac{{\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x}}{{{{\left( {1 + {{\tan }^2}x} \right)}^2} - 3{{\tan }^2}x}}} \,\,dx \\ &\qquad\qquad\qquad =\int{\frac{1+{{t}^{2}}}{1-{{t}^{2}}+{{t}^{4}}}}\,\,dt\left\{ By\text{ }the\text{ }substitution\text{ }tanx=t \right\} \end{align}\]

This is of the same form as (1) above.

If you observe carefully, you will realise that the two approaches described are more or less equivalent. Only the initial sequences of manipulations (upto (1)) are different.

**Example – 33 **

For any natural number *m*, evaluate

\(\int {\left( {{x^{3m}} + {x^{2m}} + {x^m}} \right)} \,{\left( {2{x^{2m}} + 3{x^m} + 6} \right)^{1/m}}dx,\,\,x > 0\)

**Solution: ** This expression might seem very complicated at first. But if you observe it more carefully, you will intuitively feel that the powers and coefficients of * x * in the two brackets are so adjusted such that one bracket could be expressed as a derivative of the other. It now remains to determine which of the two brackets can be expressed as a derivative of the other.

Observe that no matter what you do, you cannot express the second bracket \(\left\{ {{{\left( {2{x^{2m}} + 3{x^m} + 6} \right)}^{1/m}}} \right\}\) as a derivative of the bracket (or some part of the first bracket). Therefore, we must manipulate the expressions so that the first bracket is expressible as a derivative of the expression *inside* the second bracket.

Now observe the powers and coefficients of *x* in the second bracket carefully. If we differentiate the expression inside the second bracket

\[\left( {2{x^{2m}} + 3{x^m} + 6} \right)\]

we get

\[\left( {4m{x^{2m - 1}} + 3m{x^{m - 1}}} \right)\]

which is not what we want. In the first bracket, the coefficients of *x* are equal. Thus we want the expression inside the second bracket to be such that when we differentiate we get equal coefficients.

This is how we do it:

\[\begin{align}& I = \int {\left( {{x^{3m}} + {x^{2m}} + {x^m}} \right)} {\left( {2{x^{2m}} + 3{x^m} + 6} \right)^{1/m}}dx\\ &= \int {\left( {\frac{{{x^{3m}} + {x^{2m}} + {x^m}}}{x}} \right)} \cdot x{\left( {2{x^{2m}} + 3{x^m} + 6} \right)^{1/m}}dx\left\{ \begin{array}{l}{\rm{Multiplication\;and\;}}\\{\rm{division\;by }}x\end{array} \right\}\\& = \int {\left( {{x^{3m - 1}} + {x^{2m - 1}} + {x^{m - 1}}} \right)} \cdot {\left( {2{x^{3m}} + 3{x^{2m}} + 6{x^m}} \right)^{1/m}}dx\end{align}\]

Now, if we differentiate the expression inside the second bracket, i.e

\[2{x^{3m}} + 3{x^{2m}} + 6{x^m}\]

we get

\[\begin{array}{l}6m{x^{3m - 1}} + 6m{x^{2m - 1}} + 6m{x^{m - 1}}\\ = 6m\left( {{x^{3m - 1}} + {x^{2m - 1}} + {x^{m - 1}}} \right)\end{array}\]

This is a multiple of the first bracket! A simple manipulation (multiplication and division by *x*) thus led us to a form of the expression we wanted.

We now substitute \(2{x^{3m}} + 3{x^{2m}} + 6{x^m} = t\)

Thus, \({\left( {{x^{3m - 1}} + {x^{2m - 1}} + {x^{m - 1}}} \right)^{dx}} = \frac{{dt}}{{6m}}\)

\[\begin{align}& \Rightarrow \quad I = \frac{1}{{6m}}\int {{t^{1/m}}} \,\,dt\\ & \quad\qquad= \frac{1}{{6m}}\,\,\,\frac{{{t^{\frac{1}{m} + 1}}}}{{\frac{1}{m} + 1}}\,\, + C\\ &\quad\qquad = \frac{1}{{6\left( {m + 1} \right)}}{t^{\left( {\frac{{m + 1}}{m}} \right)}} + C\\ &\quad\qquad = \frac{1}{{6\left( {m + 1} \right)}}{\left( {2{x^{3m}} + 3{x^{2m}} + 6{x^m}} \right)^{\frac{{m + 1}}{m}}} + C\end{align}\]

This example should again prove to you that memorization is not of much use in integration!