Advanced Examples On Integration Set-3

Example - 54

Evaluate \(\begin{align}\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^{3/4}}{{\left( {x + 2} \right)}^{5/4}}}}}\end{align}\)

Solution: Observe the exponents carefully; \(\begin{align}\frac{3}{4}\end{align}\) is “ \(\begin{align}\frac{1}{4} - {\rm{less}}\end{align}\) ” than 1 while \(\begin{align}\frac{5}{4}\,\,\,\text{is}\,\,\,''\frac{1}{4}-\text{more}\end{align}\) '' than 1. Therefore, we can write the denominator as

\[{\left( {x - 1} \right)^{3/4}}{\left( {x + 2} \right)^{5/4}} = \left( {x - 1} \right)\left( {x + 2} \right){\left( {\frac{{x + 2}}{{x - 1}}} \right)^{1/4}}\]

We thus use the substitution

\[\frac{{x + 2}}{{x - 1}} = {t^4}\]

\[ \Rightarrow x = \frac{{{t^4} + 2}}{{{t^4} - 1}}\]

\[and \;dx = \frac{{ - 12{t^3}}}{{{{\left( {{t^4} - 1} \right)}^2}}}dt\]

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

\[\begin{align}&= \int {\frac{{ - 12{t^3}dt}}{{\left( {\frac{3}{{{t^4} - 1}}} \right)\left( {\frac{{3{t^4}}}{{{t^4} - 1}}} \right) \cdot t \cdot {{\left( {{t^4} - 1} \right)}^2}}}} \\& = - \frac{4}{3}\int {\frac{{dt}}{{{t^2}}}} \\
&= \frac{4}{{3t}} + C\\&= \frac{4}{3}{\left( {\frac{{x - 1}}{{x + 2}}} \right)^{1/4}} + C\end{align}\]

Example - 55

Evaluate \(\begin{align}\int {\left( {x - 2} \right)\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx\end{align}\)

Solution: \[I = \int {\left( {x - 1} \right)\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx - \int {\sqrt {\frac{{1 + x}}{{1 - x}}} } \,\,dx\]

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

The first integral is of the standard form (27) To evaluate the second integral I₁ , we can of course use the substitution \(x = \cos 2\theta .\) However, here we use the substitution described in the preceeding discussion.

\[\begin{align} &\quad\quad\frac{1+x}{1-x}={{t}^{2}}  \\&\Rightarrow \quad  x=\frac{{{t}^{2}}-1}{{{t}^{2}}+1}  \\&\Rightarrow \quad dx=\frac{4t}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt  \\ &\Rightarrow \quad  {{I}_{1}}=\int{t\cdot \frac{4tdt}{{{\left( {{t}^{2}}+1 \right)}^{2}}}}  \\&\qquad\quad\;  =4\int{\frac{{{t}^{2}}}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt}  \\&\qquad\quad\;=4\left\{ \int{\frac{1}{{{t}^{2}}+1}dt-\int{\frac{1}{{{\left( {{t}^{2}}+1 \right)}^{2}}}dt}} \right\}  \\&\qquad\qquad\qquad \left( \begin{gathered} \text{substitute}\;t=\tan \theta   \\\Rightarrow dt={{\sec }^{2}}\theta d\theta \end{gathered} \right)  \\ &\qquad\quad\;=4\left\{ {{\tan }^{-1}}t-\int{{{\cos }^{2}}\theta d\theta } \right\}  \\&\qquad\quad\; =4\left\{ {{\tan }^{-1}}t-\frac{1}{2}{{\tan }^{-1}}t-\frac{1}{4}\left( \frac{2t}{1+{{t}^{2}}} \right) \right\}+C  \\&\qquad\quad\;=2{{\tan }^{-1}}t-\frac{2t}{1+{{t}^{2}}}+C  \end{align}\]

This example once more shows that it is not necessary that there must be only one (unique) substitution for a given integral. Multiple possible substitutions exist for many integrals.