Advanced Examples On Integration Set-5

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Example - 57

Evaluate the following integrals:

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

(b) \(\begin{align}\int {\sqrt {2 + {{\tan }^2}x} } \,\,dx\end{align}\)

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

(d) \(\begin{align}\int {\frac{x}{{\left( {{x^2} + 4} \right)\sqrt {{x^2} + 9} }}} \,\,dx\end{align}\)

Solution: (a) The denominator has quadratic factors which are “purely quadratic”. Thus we use the substitution \(x = \frac{1}{t}\) (and see what happens):

\[\begin{align}&\qquad\quad x=\frac{1}{t} \\& \Rightarrow \quad dx=-\frac{1}{{{t}^{2}}}dt \\ 
 & I=\int{\frac{1}{{{x}^{2}}\sqrt{1+{{x}^{2}}}}}\,\,dx \\ &\;\; =\int{\frac{{{t}^{2}}}{\sqrt{1+\frac{1}{{{t}^{2}}}}}}\,\cdot -\frac{1}{{{t}^{2}}}\,dt \\ & \;\;=\int{\frac{t}{\sqrt{{{t}^{2}}+1}}}\,dt \\ 
 & \;\;=-\int{dz}\,\,\,\text{(using the substitution }{{\text{t}}^{\text{2}}}\text{ + 1 =  }{{\text{z}}^{\text{2}}}\text{)} \\ & \;\;=-z+C \\ &\;\; =-\sqrt{{{t}^{2}}+1}+C \\ &\;\; =-\frac{\sqrt{1+{{x}^{2}}}}{x}+C  \end{align}\]

(b) Substitute 

\[\begin{align}&\qquad 2+{{\tan }^{2}}x={{z}^{2}} \\ & \Rightarrow\quad 2\tan x{{\sec }^{2}}xdx=2zdz \\ 
 & \Rightarrow\quad I=\int{\sqrt{2+{{\tan }^{2}}x}}\,\,dx \\ 
 & \qquad\quad=\int{z\cdot \frac{2zdz}{2\tan x{{\sec }^{2}}x}}\, \\ 
 &\qquad\quad =\int{\frac{{{z}^{2}}}{\sqrt{{{z}^{2}}-2}\left( {{z}^{2}}-1 \right)}}\,dz \\ 
 &\qquad\quad =\int{\frac{\left( {{z}^{2}}-1 \right)+1}{\left( {{z}^{2}}-1 \right)\sqrt{{{z}^{2}}-2}}}\,dz \\ 
 &\qquad\quad =\int{\frac{1}{\sqrt{{{z}^{2}}-2}}}\,dz+\int{\frac{1}{\left( {{z}^{2}}-1 \right)\sqrt{{{z}^{2}}-2}}}\,\,dz \\  & \qquad\quad=\ln \left| z+\sqrt{{{z}^{2}}-1} \right|+{{I}_{1}} \\ 
\end{align}\]

In I1 , both the quadratic factors are purely quadratic, so we use the substitution \(\begin{align}z = \frac{1}{y}\end{align}\)

\[\begin{align}& \Rightarrow\quad  dz=-\frac{1}{{{y}^{2}}}dy \\ 
 & \Rightarrow \quad {{I}_{1}}=\int{\frac{1}{\left( {{z}^{2}}-1 \right)\sqrt{{{z}^{2}}-2}}}\,\,dz \\ 
 &\qquad\quad =\int{\frac{{{y}^{2}}}{\left( 1-{{y}^{2}} \right)\frac{\sqrt{1-2{{y}^{2}}}}{y}}}\cdot \,-\frac{1}{{{y}^{2}}}dy \\ & \qquad\quad=\int{\frac{-y}{\left( 1-{{y}^{2}} \right)\sqrt{1-2{{y}^{2}}}}}\,\,dy \\ 
\end{align}\]

This can now be simply evaluated using the substitution \(1 - 2{y^2} = {u^2},\) so that

\[\begin{align}
  &\qquad -ydy=\frac{du}{2} \\ 
 & \Rightarrow\quad {{I}_{1}}=\frac{1}{2}\int{\frac{udu}{\frac{\left( {{u}^{2}}+1 \right)}{2}\cdot u}} \\ 
 &\qquad \quad f =\int{\frac{du}{{{u}^{2}}+1}} \\ \\
 & \qquad \quad\;={{\tan }^{-1}}u+C \\ 
 & \qquad \quad\;={{\tan }^{-1}}\left( \sqrt{1-2{{y}^{2}}} \right)+C \\ 
 & \qquad \quad\;={{\tan }^{-1}}\left( \frac{\sqrt{{{z}^{2}}-2}}{z} \right)+C \\ 
 & \qquad \quad\;={{\tan }^{-1}}\left( \frac{\tan x}{\sqrt{2+{{\tan }^{2}}x}} \right)+C \\ 
\end{align}\]

Thus, using I1 , I  becomes known.

(c) Observe that the quadratic factor \({x^2} + 3x + 2\) can be factorised, so that we can write:

\[\begin{align}& I=\int{\frac{1}{\left( {{x}^{2}}+3x+2 \right)\sqrt{{{x}^{2}}-1}}}\,\,dx \\ &\; =\int{\frac{1}{\left( x+1 \right)\left( x+2 \right)\sqrt{{{x}^{2}}-1}}}\,\,dx \\ & \;=\int{\left( \frac{1}{x+1}-\frac{1}{x+2} \right)\frac{1}{\sqrt{{{x}^{2}}-1}}}\,\,dx \\ & \;=\int{\frac{1}{\left( x+1 \right)\sqrt{{{x}^{2}}-1}}}\,\,dx-\int{\frac{1}{\left( x+2 \right)\sqrt{{{x}^{2}}-1}}}\,\,dx \\ & \;={{I}_{1~}}-\text{ }{{I}_{2}} \\ 
\end{align}\]

Calculating I1 and I2 separately is much simpler than calculating I.

For I1 , we substitute \(\begin{align}x + 1 = \frac{1}{t}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,dx = - \frac{1}{{{t^2}}}dt\end{align}\)

\[ \Rightarrow \quad {I_1} = \int {\frac{{ - dt}}{{\sqrt {1 - 2t} }}} \]

\[\begin{align}& = \sqrt {1 - 2t} \\& = \sqrt {\frac{{x - 1}}{{x + 1}}} \end{align}\]

For I2 , we substitute     \(\begin{align}x + 2 = \frac{1}{t}\,\,\,\, \Rightarrow \,\,\,\,dx = - \frac{1}{{{t^2}}}dt\,\end{align}\)

\[\begin{align}& \Rightarrow\quad {{I}_{2}}=-\int{\frac{dt}{\sqrt{1-4t+3{{t}^{2}}}}} \\ 
 &\qquad\quad\;=-\frac{1}{\sqrt{3}}\ln \left| \left( t-\frac{2}{3} \right)+\sqrt{{{t}^{2}}-\frac{4}{3}t+\frac{1}{3}} \right|\left( \text{Verify} \right)  \\\\& \qquad\quad\; =-\frac{1}{\sqrt{3}}\ln \left| \frac{-\left( 2x+1 \right)}{3\left( x+2 \right)}+\sqrt{\frac{{{x}^{2}}-1}{3{{\left( x+2 \right)}^{2}}}} \right|  \end{align}\]

From I1 and I2 , I becomes known.

(d) This integral can be evaluated using the simple substitution \({x^2} + 9 = {z^2}\, \Rightarrow \,\,xdx = zdz\)

\[\begin{align}& \Rightarrow \quad I=\int{\frac{x}{\left( {{x}^{2}}+4 \right)\sqrt{{{x}^{2}}+9}}}dx \\ 
 & \qquad\quad=\int{\frac{z}{\left( {{z}^{2}}-5 \right)\cdot z}}dz  \\   {}  \\&\qquad\quad=\frac{1}{2\sqrt{5}}\ln \left| \frac{z-\sqrt{5}}{z+\sqrt{5}} \right|+C  \\ {}  \\&\qquad\quad=\frac{1}{2\sqrt{5}}\ln \left| \frac{\sqrt{{{x}^{2}}+9}-\sqrt{5}}{\sqrt{{{x}^{2}}+9}+\sqrt{5}} \right|+C  \end{align}\]

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