ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 6- Worked Out Examples

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

	$\sqrt x = \cos \theta$
	$\Rightarrow \,\,\,\,x = {\cos ^2}\theta$
	$\Rightarrow\,\,\,\, dx = - 2\sin \theta \cos \theta d\theta$

	$I = \int {\sqrt {\dfrac{{1 - \sqrt x }}{{1 + \sqrt x }}} .\dfrac{1}{x}dx} = \int {\sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} \dfrac{1}{{{{\cos }^2}\theta }}.\left( { - 2\sin \theta \cos \theta d\theta } \right)}$
	$= - 2\int {\sqrt {\dfrac{{2{{\sin }^2}\left( {\theta /2} \right)}}{{2{{\cos }^2}\left( {\theta /2} \right)}}} \tan \theta d\theta }$
	$= - 2\int {\tan \left( {\dfrac{\theta }{2}} \right).\tan \theta } d\theta$
	$= - 4\int {\dfrac{{{{\sin }^2}\left( {\theta /2} \right)}}{{\cos \theta }}} d\theta$
	$= - 2\int {\dfrac{{1 - \cos \theta }}{{\cos \theta }}} d\theta$
	$= - 2\int {\left( {\sec \theta - 1} \right)} d\theta$
	$= - 2\left\{ {\ln \left\| {\sec \theta + \tan \theta } \right\| - \theta } \right\} + C$
	$= - 2\ln \left\| {\dfrac{{1 + \sin \theta }}{{\cos \theta }}} \right\| + 2\theta + C$
	$= - 2\ln \left( {\dfrac{{1 + \sqrt {1 - x} }}{{\sqrt x }}} \right) + 2{\cos ^{ - 1}}\sqrt x + C$

Monday, 4 August 2014

CHAPTER 6- Worked Out Examples

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