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Examples On Integration By Substitution Set-8

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Example – 23

Evaluate the following integrals

(a) \(\begin{align}\int {\frac{{{x^2} + 1}}{{{x^4} + 1}}\,\,dx} \end{align}\) (b) \(\begin{align}\int {\frac{{{x^2} - 1}}{{{x^4} + 1}}\,\,dx} \end{align}\) (c) \(\begin{align}\int {\frac{{{x^2} + 1}}{{{x^4} + {x^2} + 1}}\,\,dx} \end{align}\) (d) \(\begin{align}\int {\frac{1}{{{x^4} + 1}}\,\,dx} \end{align}\)

Solution: (a) \(\begin{align} I = \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}\,\,dx} \end{align}\)

A slight algebraic manipulation of the given expression will lead us to a possible substitution; we divide the numerator and denominator by \({x^2}\) .

\[I = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}\,\,dx} \]

We now need to express the numerator as a differential of some term occurring in the denominator. The denominator \(\begin{align}{x^2} + \frac{1}{{{x^2}}}\end{align}\) can be written \(\begin{align}{\left( {x - \frac{1}{x}} \right)^2} + 2\end{align}\) . The derivative of \(\begin{align}x - \frac{1}{x}{\rm{ }}\;is\;{\rm{ }}x + \frac{1}{{{x^2}}}\end{align}\) . Thus, we substitute \(\begin{align} x - \frac{1}{x} = t\end{align}\) . The integral reduces to \(\begin{align} I = \int {\frac{{dt}}{{{t^2} + 2}}} \end{align}\) .

This is of the form encountered in Example-18 part-b. (Refer to that example for elaboration).

The value of the integral is

\[I = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} \right) + C\\ \quad\;\;\;= \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + C\]
 

(b) \(\begin{align} I = \int {\frac{{{x^2} - 1}}{{{x^4} + 1}}\,\,dx} \end{align}\)

We follow the same approach as we did in the previous part:

\[\begin{align} I = \int {\frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}\,\,dx} \end{align}\]

However, this time we write the denominator \(\begin{align}\left( {{x^2} + \frac{1}{{{x^2}}}} \right)as{\left( {x + \frac{1}{x}} \right)^2} - 2\end{align}\) and substitute \(\begin{align}\left( {x + \frac{1}{x}} \right) = t\end{align}\) ; this is because the numerator \(\begin{align}\left( {1 - \frac{1}{{{x^2}}}} \right)\end{align}\) , is a derivative of \(\begin{align}\left( {x + \frac{1}{x}} \right)\end{align}\) , while in the previous example, the numerator was \(\begin{align}\left( {1 + \frac{1}{{{x^2}}}} \right)\end{align}\) , which was a derivative of \(\begin{align}\left( {x - \frac{1}{x}} \right)\end{align}\) .

Thus, we have,

\[\begin{align}& x + \frac{1}{x} \;= t\\ &\Rightarrow\quad I = \int {\frac{{dt}}{{{t^2} - 2}}} \end{align}\]
 

This is of the form encountered in Example - 18 part (d)

\[ \Rightarrow\quad I = \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{t - \sqrt 2 }}{{t + \sqrt 2 }}} \right| + C\\ \qquad\qquad\qquad= \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{{x^2} - \sqrt 2 x + 1}}{{{x^2} + \sqrt 2 x + 1}}} \right| + C\]

(c)\( \begin{align} I = \int {\frac{{{x^2} + 1}}{{{x^4} + {x^2} + 1}}\,\,dx} = \int {\frac{{1 + \frac{1}{{{x^2}}}\,\,}}{{{x^2} + \frac{1}{{{x^2}}} + 1}}} \,\,dx \end{align}\)

\(\qquad\qquad\qquad\qquad\begin{align}\\ \qquad= \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 3}}dx}\end{align}\)

The substitution \(\begin{align}x - \frac{1}{x} = t\end{align}\) reduces \(I\) to an “integrable” form:

\[\begin{align}&I = \int {\frac{{dt}}{{{t^2} + 3}}}\\& \;\;= \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right) + C\,\\ &\;\; = \frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 3 x}}} \right) + C\end{align}\]

(d) \(\begin{align}I = \int {\frac{1}{{{x^4} + 1}}\,\,dx} \end{align}\)

This integral can easily be obtained from the integrals we’ve already evaluated in parts (a) and (b):

\[\,\,\,\,\,I = \int {\frac{1}{{{x^4} + 1}}\,\,dx} = \frac{1}{2}\left\{ {\int {\frac{{{x^2} + 1}}{{{x^4} + 1}}dx - \int {\frac{{{x^2} - 1}}{{{x^4} + 1}}dx} } } \right\}\]

Example - 24

Evaluate \(\int {\sqrt {\tan \theta } d\theta } \) .

Solution: To get rid of the square-root and convert this expression into a purely rational one, we use the substitution

\[\tan \theta = {x^2}\]

Thus,                                                                                           \({\sec ^2}\theta \,d\theta = 2x\,\,dx\)

\[\begin{align}&\Rightarrow\quad d\theta = \frac{{2x\,\,dx}}{{{{\sec }^2}\theta }}\\&\quad\qquad\;\,= \frac{{2x}}{{1 + {x^4}}}\,\,dx\\&\Rightarrow\quad I = \int {\sqrt {{x^2}} } \cdot \frac{{2x}}{{1 + {x^4}}}\,\,dx\\& \qquad\quad= 2\int {\frac{{{x^2}}}{{1 + {x^4}}}\,\,dx} \\&\quad\qquad= \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}\,\,dx} + \int {\frac{{{x^2} - 1}}{{{x^4} + 1}}\,\,dx} \end{align}\]

We have already evaluated both these integrals in the previous examples.

Example - 25

Evaluate \(\begin{align}\int {\frac{{{x^8}}}{{{x^{12}} - 1}}\,\,dx} \end{align}\)

Solution : To simplify this integral, we need to make an appropriate substitution, such as \({x^n} = t\) where \(n < 12\) , n should of course be a factor of 12 and the numerator \(({x^8}\,\,dx)\) should be expressible in terms of t and the differential of t. A little though will show that \({x^3} = t\) is the appropriate substitution possible here:

\[\begin{align}&\quad\quad{x^3} = t\\ &\Rightarrow \,\,\,\,\,3{x^2}\,\,dx = dt\\ &\Rightarrow \quad I = \int {\frac{{{x^8}}}{{{x^{12}} - 1}}\,\,dx}\\ &\qquad\quad = \int {\frac{{{{({x^3})}^2} \cdot {x^2}}}{{{{({x^3})}^4} - 1}}\,\,dx} \\ &\qquad\quad = \frac{1}{3}\int {\frac{{{t^2}}}{{{t^4} - 1}}\,\,dt} \left( \begin{array}{l}{\rm{The\;numerator\;was\;written\;like\;this\;so}}\\{\rm{\;that\;the\;two\;factors\;in\;the\;denominator\;}}\\{\rm{get\;separated\;into\;different \;fractions}}\end{array} \right)\\ &\qquad\quad= \frac{1}{3}\int {\frac{{{t^2}}}{{({t^2} - 1)({t^2} + 1)}}\,\,dt} \\ &\qquad\quad= \frac{1}{6}\int {\frac{{({t^2} - 1) + ({t^2} + 1)}}{{({t^2} - 1)({t^2} + 1)}}\,\,dt} \\ &\qquad\quad{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\left\{ {\int {\frac{1}{{{t^2} + 1}}\,\,dt + \int {\frac{1}{{{t^2} - 1}}\,\,dt} } } \right\}\\ &\qquad\quad{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}\left( {{{\tan }^{ - 1}}t + \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right) + C\\ &\qquad\quad{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}{\tan ^{ - 1}}{x^3} + \frac{1}{{12}}\ln \left| {\frac{{{x^3} - 1}}{{{x^3} + 1}}} \right| + C\end{align}\]

In the preceding examples, we have seem a lot many integrals, which could be evaluated using the general results derived in Example-18. In fact, these results are very important and many integrals that we’ll encounter can be reduced by appropriate manipulations to one of these general forms.

The following table extends Table-1 on Page-2 and includes these general forms as “standard integral forms”. (This table is being presented here for your convenience and easy reference-the results are from Example-18)

 

Table - 1 Continued

Standard integrals

 Integrals   Substitution used   Result
15. \(\begin{align}\int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}\,\,dx} \end{align}\) \(x = a\sin \theta \) \(\begin{align}{\sin ^{ - 1}}\frac{x}{a} + C\end{align}\)
17. \(\begin{align}\int {\frac{1}{{{a^2} + {x^2}}}\,\,dx} \end{align}\) \(x = a\tan \theta \) \(\frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} + C\)
19. \(\begin{align}\int {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}\,\,dx} \end{align}\) \(x = a\sec \theta \) \(\frac{1}{a}{\sec ^{ - 1}}\frac{x}{a} + C\) 
21. \(\begin{align}\int {\frac{1}{{{x^2} - {a^2}}}\,\,dx} \end{align}\) \(\begin{array}{l} x = a\sec \theta \\ \left( \begin{array}{l}{\rm{Alternatively, the expression}}\\{\rm{can be split in to separate fractions}}\end{array} \right)\end{array}\) \(\frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C\)
22. \(\begin{align}\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}\,\,dx} \end{align}\) \(x = a\tan \theta \) \(\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\)
23. \(\begin{align}\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}\,\,dx} \end{align}\) \(x = a\sec\theta \) \(\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C\)
24.* \(\begin{align}\int {\sqrt {\frac{{a - x}}{{a + x}}} \,\,dx} \end{align}\) \(x = a\cos 2\theta \) \(\sqrt {{a^2} - {x^2}} - a{\cos ^{ - 1}}\frac{x}{a} + C\)
25.* \(\begin{align}\int {\sqrt {\frac{{x - a}}{{b - x}}} \,\,dx} \end{align}\) \(x = a{\cos ^2}\theta + b{\sin ^2}\theta \) \[\left( {b - a} \right)\left\{ {\frac{1}{2}{{\cos }^{ - 1}}\frac{x}{a} - \frac{1}{{2a}}\sqrt {{a^2} - {x^2}} } \right\} + C\]

* need not be memorized

We will extend this table further as we progress through the rest of this chapter. As stated earlier, many integrals can be converted to these ‘standard’ forms and thats where lies the use in committing these results to memory. We saw in example-19 some questions based on these general forms. We will now discuss in more detail how to convert a given integral into one of these standard forms, if possible.

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Download SOLVED Practice Questions of Examples On Integration By Substitution Set-8 for FREE
Indefinite Integration
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