Advanced Examples On Integration Set-6

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We now turn our attention to a special class of irrational algebraic functions, which are, though rarely encountered in any exam, are discussed here for the sake of a comprehensive coverage.

These are integrals of the form \(R\left( {x,\,\,\sqrt {Q(x)} } \right)\) . For example, an integral like \(\begin{align}\int {\frac{{dx}}{{x + \sqrt {{x^2} - x + 1} }}} \end{align}\) would fall in this class. Note that uptill now we have not encountered integrals of this form.

A specific substitution scheme described below is used to solve such integrals.

Let \(Q(x) = a{x^2} + bx + c\)

\(\Rightarrow \) If a > 0, substitute \(\sqrt {a{x^2} + bx + c} = t \pm x\sqrt a \) .

i.e. \(a{x^2} + bx + c = {t^2} + a{x^2} - 2\sqrt a \,tx\)

\(\Rightarrow \) If c > 0, substitute \(\sqrt {a{x^2} + bx + c} = tx \pm \sqrt c \)

i.e. \(a{x^2} + bx + c = {t^2}{x^2} + {c^2} - 2\sqrt c \,\,tx\)

\(\Rightarrow \) If \(a{x^2} + bx + c\) is factorisable, i.e. it has real roots

\(\alpha \,\,{\rm{and }}\beta \) such that \(a{x^2} + bx + c = a(x - \alpha )(x - \beta )\) ,

Substitute \(\sqrt {a{x^2} + bx + c} = (x - \alpha )t\) .

These substitutions are known as Euler’s substitutions. You will get an idea of why these substitutions work from the following example.

Example – 58

Evaluate the following integrals:

(a) \(\begin{align}\int {\frac{1}{{x + \sqrt {{x^2} - x + 1} }}} \,\,dx\end{align}\)

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

Solution: (a) From the Euler substitution scheme, we use the substitution \(\sqrt {{x^2} - x + 1} = t \pm x\)

\(\Rightarrow \) \({x^2} - x + 1 = {t^2} + {x^2} \pm 2tx\)

\(\Rightarrow \) \((1 \pm 2t)x = 1 - {t^2}\)

\(\Rightarrow \) \(\begin{align}x = \frac{{1 - {t^2}}}{{1 \pm 2t}}\end{align}\)

Let us select the “–” sign. Thus,

\[\begin{align} & dx=\frac{(1-2t)(-2t)-(1-{{t}^{2}})(-2)}{{{(1-2t)}^{2}}}dt \\  &\quad =\frac{3+2{{t}^{2}}}{{{(1-2t)}^{2}}}dt \end{align}\]

Also, 

\(\begin{align}&x + \sqrt {{x^2} - x + 1} = x + t - x\\\\ &\qquad\qquad\qquad\quad= t\end{align}\)

Thus, \(\begin{align}I = \int {\frac{1}{t} \cdot \frac{{3 + 2{t^2}}}{{{{(1 - 2t)}^2}}}\,\,dt} \end{align}\)

\(\, = \int {\frac{{3 + 2{t^2}}}{{t{{(1 - 2t)}^2}}}\,\,dt} \)

It is obvious that this integral can now be solved using expansion by partial fractions.

You should now understand why the Euler substitutions work. They reduce the integral to a rational from which can be expanded using partial fractions.

(b) The Euler substitution scheme tells us that we can use the following substitution.

\[\begin{align}&\qquad\quad \sqrt{{{x}^{2}}+2x+4}=t+x\left( \begin{gathered}
  & \text{We took the positive sign } \\ 
 & \text{since then the denominator } \\ 
 & x-\sqrt{{{x}^{2}}+2x+4}\text{ simply} \\ 
 & \text{becomes -}t \\ 
\end{gathered} \right) \\ 
 & \Rightarrow \quad {{x}^{2}}+2x+4={{t}^{2}}+{{x}^{2}}+2tx \\ 
 & \Rightarrow \quad x=\frac{4-{{t}^{2}}}{2t-2} \\ 
 & \Rightarrow \quad dx=\frac{1}{2}\frac{(t-1)(-2t)-(4-{{t}^{2}})(1)}{{{(t-1)}^{2}}}\,\,dt \\ 
 & \qquad\qquad=\frac{1}{2}\frac{-{{t}^{2}}+2t-4}{{{(t-1)}^{2}}}\,\,dt \\ 
 & \qquad\qquad=\frac{-({{t}^{2}}-2t+4)}{2\,{{(t-1)}^{2}}}\,\,dt \\ 
 & \Rightarrow \quad I=\int{\frac{1}{-t}\cdot \frac{-({{t}^{2}}-2t+4)}{2{{(t-1)}^{2}}}\,\,dt} \\ 
 & \qquad\;\;\;=\int{\frac{{{t}^{2}}-2t+4}{2t{{(t-1)}^{2}}}\,\,dt} \\ 
 &\qquad\;\;\;=\int{\frac{{{(t-1)}^{2}}+3}{2t{{(t-1)}^{2}}}\,\,dt} \\ &
 \qquad\;\;\;=\int{\frac{dt}{2t}+\frac{3}{2}\int{\frac{1}{t{{(t-1)}^{2}}}}\,\,dt} \\ 
\end{align}\]

The second integral can easily be evaluated using expansion by partial fractions.

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Download SOLVED Practice Questions of Advanced Examples On Integration Set-6 for FREE
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