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Monday, 4 August 2014

CHAPTER 6- Worked Out Examples

    Example: 1     

Evaluate \int {\dfrac{1}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}}dx} .
Solution: 1

I = \int {\dfrac{1}{{{{\left( {{x^2} + 2x + 2} \right)}^2}}}dx}
 = \int {\dfrac{1}{{{{\left( {{{\left( {x + 1} \right)}^2} + 1} \right)}^2}}}dx}
The denominator can be reduced by the substitution
x + 1 = \tan \theta
 \Rightarrow  \,\,\,\, {\left( {x + 1} \right)^2} + 1 = {\tan ^2}\theta  + 1 = {\sec ^2}\theta
Also,
dx = {\sec ^2}\theta d\theta
 \Rightarrow\,\,\,\, I = \int {\dfrac{{{{\sec }^2}\theta }}{{{{\sec }^4}\theta }}d\theta }
 = \int {{{\cos }^2}\theta d\theta }
 = \dfrac{1}{2}\int {\left( {1 + \cos 2\theta } \right)d\theta }
 = \dfrac{1}{2}\left( {\theta  + \dfrac{{\sin 2\theta }}{2}} \right) + C
 = \dfrac{1}{2}\left( {\theta  + \sin \theta \cos \theta } \right) + C
To express the integral in terms of x, we use
x + 1 = \tan \theta
\, \Rightarrow\,\,\,\, \theta  = {\tan ^{ - 1}}\left( {x + 1} \right)
 \Rightarrow\,\,\,\, \sin \theta  = \dfrac{{\left( {x + 1} \right)}}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }}
and \,\,\,\,\,\,\,\, \cos \theta  = \dfrac{1}{{\sqrt {1 + {{\left( {x + 1} \right)}^2}} }}
 \Rightarrow I = \dfrac{1}{2}\left\{ {{{\tan }^{ - 1}}\left( {x + 1} \right) + \dfrac{{x + 1}}{{{x^2} + 2x + 2}}} \right\} + C
     Example: 2      

Evaluate \int {\sqrt {\dfrac{{1 - \sqrt x }}{{1 + \sqrt x }}.} \dfrac{1}{x}dx}
Solution: 2

A slight thought on the form of this expression will hint that a trignometric substitution might help; recall that both \left( {1 - \cos \theta } \right) and \left( {1 + \cos \theta } \right) can be reduced to single terms.Therefore, we use the substitution
\sqrt x  = \cos \theta
 \Rightarrow \,\,\,\,x = {\cos ^2}\theta
 \Rightarrow\,\,\,\, dx =  - 2\sin \theta \cos \theta d\theta
I = \int {\sqrt {\dfrac{{1 - \sqrt x }}{{1 + \sqrt x }}} .\dfrac{1}{x}dx}  = \int {\sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} \dfrac{1}{{{{\cos }^2}\theta }}.\left( { - 2\sin \theta \cos \theta d\theta } \right)}
 =  - 2\int {\sqrt {\dfrac{{2{{\sin }^2}\left( {\theta /2} \right)}}{{2{{\cos }^2}\left( {\theta /2} \right)}}} \tan \theta d\theta }
 =  - 2\int {\tan \left( {\dfrac{\theta }{2}} \right).\tan \theta } d\theta
 =  - 4\int {\dfrac{{{{\sin }^2}\left( {\theta /2} \right)}}{{\cos \theta }}} d\theta
 =  - 2\int {\dfrac{{1 - \cos \theta }}{{\cos \theta }}} d\theta
 =  - 2\int {\left( {\sec \theta  - 1} \right)} d\theta
 =  - 2\left\{ {\ln \left| {\sec \theta  + \tan \theta } \right| - \theta } \right\} + C
 =  - 2\ln \left| {\dfrac{{1 + \sin \theta }}{{\cos \theta }}} \right| + 2\theta  + C
 =  - 2\ln \left( {\dfrac{{1 + \sqrt {1 - x} }}{{\sqrt x }}} \right) + 2{\cos ^{ - 1}}\sqrt x  + C
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