Examples On Integration By Substitution Set-9

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Suppose that we have a quadratic expression \(Q(x) = a{x^2} + bx + c\). It should be obvious that integrals of the form \(\begin{align}\int {\frac{1}{{Q(x)}}} \,\,dx\end{align}\) can be evaluated using either (17) or (21) {depending on whether Q(x) takes the form of \({X^2} + {A^2}\) or \({A^2} - {X^2}\) upon rearrangement ; X depends on x linearly; A is a constant}. Similarly, integrals of the form \(\begin{align}\int {\frac{1}{{\sqrt {Q(x)} }}} \,\,dx\end{align}\) can be evaluate using (15), (22), or (23) depending on what form Q(x) takes upon rearrangement. Let us go through some examples related to these forms.

Example - 26 Evaluate the following integrals:

(a) \(\begin{align}\int {\frac{1}{{{x^2} + 3x + 2}}} \,\,dt\end{align}\) (b)\(\begin{align}\int {\frac{1}{{4{x^2} + 4x + 2}}} \,\,dt\end{align}\)
(c) \(\begin{align}\int {\frac{1}{{\sqrt {4{x^2} + 5x + 4} }}} \,\,dt\end{align}\) (d) \(\begin{align}\int {\frac{1}{{\sqrt { - 4{x^2} - 4x} }}} \,\,dt\end{align}\)

Solution: Our attempts in each of these questions will be to reduce (rearrange) the quadratic expressions so that it resembles one of our standard forms of Table - 1

(a)                                                                                          \(\begin{align}{x^2} + 3x + 2 = {\left( {x + \frac{3}{2}} \right)^2} - \frac{1}{4}\end{align}\)

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

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

This is of the form (21) where \(\begin{align}a = \frac{1}{2}\end{align}\)

\[I = \ln \left( {\frac{{x + 1}}{{x + 2}}} \right) + C\]

(b)                                                                                               \(4{x^2} + 4x + 2 = \left( {4{x^2} + 4x + 1} \right) + 1\)

\[ = {\left( {2x + 1} \right)^2} + {1^2}\]

Thus,                                                                                                          \(\begin{align}I = \int {\frac{1}{{{{\left( {2x + 1} \right)}^2} + {1^2}}}} \,\,dx\end{align}\)

This is of the form (17) which contains the expression \({X^2} + {A^2}\) ; \(X = 2x + 1\,\,\,{\rm{and}}\,\,A = 1\)

Thus,

\[\begin{align}&I = \frac{1}{\begin{array}{l}\,2\\   \nwarrow \end{array}}{\tan ^{ - 1}}\left( {{\rm{2}}x{\rm{ + 1}}} \right){\rm{ + }}C\\&{\text{We have to divide by the coefficient }}\\&{\text{of}}\;x\;{\text{ in}}\times \;\;{\text{;}}\,\,{\text{refer}}\,{\text{to}}\,{\text{page}}\,{\text{1}}\,\,{\text{property}} - (e)\end{align}\]

(c)                                                                             \(4{x^2} + 5x + 4 = 4\left( {{x^2} + \frac{5}{4}x + 1} \right)\)

\[\begin{align}& = 4\left( {{{\left( {x + \frac{5}{8}} \right)}^2} + 1 - {{\left( {\frac{5}{8}} \right)}^2}} \right)\\ &= 4{\left( {x + \frac{5}{8}} \right)^2} + \frac{{39}}{{16}}\\& = {\left( {2x + \frac{5}{4}} \right)^2} + {\left( {\frac{{\sqrt {39} }}{4}} \right)^2}\end{align}\]

The integral is therefore of the form (22). The result is

\[\begin{align}&I = \frac{1}{2}\ln \left| {\left( {2x + \frac{5}{4}} \right) + \sqrt {4{x^2} + 5x + 4} } \right| + C\\&\,\,\,\,\,\,\,\,\,\,\,\, \nwarrow \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{again, division by the coeff of }}x\end{align}\]

(d)                              \(\begin{align}- 4{x^2} - 4x &= 1 - \left( {4{x^2} + 4x + 1} \right)\\ &= {1^2} - {\left( {2x + 1} \right)^2}\end{align}\)

The integral is therefore of the form (15).

The result is

\[\begin{array}{l}I = \frac{1}{2}{\sin ^{ - 1}}\left( {2x + 1} \right) + C\\\,\,\,\,\,\,\,\,\,\,\,\, \nwarrow \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{again, division by the coeff of }}x\,\end{array}\]

Example - 27

Evaluate \(\int {\sqrt {\sec x - 1} } \,\,dx\)

Solution:                                                                                            \(I = \int {\sqrt {\sec x - 1} } \,\,dx\)

\[\begin{align}& = \int {\sqrt {\frac{{1 - \cos x}}{{\cos x}}} } \,\,dx\\ &= \int {\sqrt {\frac{{1 - {{\cos }^2}x}}{{\cos x\left( {1 + \cos x} \right)}}} } \,\,dx\\ &= \int {\frac{{\sin x}}{{\sqrt {{{\cos }^2}x + \cos x} }}} \,\,dx\end{align}\]

Now we use the substitution cos x = t. This will reduce the trignometric form above to an algebraic form:

\[\begin{align}&\qquad\;\; \cos x = t\\& \Rightarrow\quad \sin xdx = - dt\\ &\Rightarrow\quad I = - \int {\frac{{dt}}{{\sqrt {{t^2} + t} }}} \\&\quad\qquad= - \int {\frac{{dt}}{{\sqrt {{{\left( {t + \frac{1}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} }}} \;\;\; \;\left\{ \begin{array}{l}{\rm{Conversion\;to\;the\;}}\\{\rm{standard\;form\;(23)}}\end{array} \right\}\\ &\quad\qquad= - \ln \left| {\left( {t + \frac{1}{2}} \right) + \sqrt {{t^2} + t} } \right| + C\\&\quad\qquad = - \ln \left| {\left( {\cos x + \frac{1}{2}} \right) + \sqrt {{{\cos }^2}x + \cos x} } \right| + C\end{align}\]

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Download SOLVED Practice Questions of Examples On Integration By Substitution Set-9 for FREE
Indefinite Integration
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