ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 15- Worked Out Examples

Monday, 4 August 2014

CHAPTER 15- Worked Out Examples

Example: 1

Evaluate the following integrals:

	(a) $\int {\dfrac{{{x^2} + x + 1}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx$
	(b) $\int {\left( {{x^2} + x + 1} \right)} \,\sqrt {{x^2} + 2x + 3} \,\,dx$

Solution: 1-(a)

Let us first find the constants $\alpha ,\,\beta \,$ and $\gamma$ which will be common to both the questions.

	${x^2} + x + 1 = \alpha \left( {{x^2} + 2x + 3} \right) + \beta \left( {2x + 2} \right) + \gamma$
	$= \alpha {x^2} + \left( {2\alpha + 2\beta } \right)x + 3\alpha + 2\beta + \gamma$
	$\Rightarrow\,\,\,\, \alpha = 1,\,\,\,\beta = \,\, - \dfrac{1}{2},\,\,\,\,\,\,\,\gamma = - 1$
	${I_1} = \int {\dfrac{{{x^2} + x + 1}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx$
	$= \int {\dfrac{{{x^2} + 2x + 3}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx - \dfrac{1}{2}\int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx - \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx$
	$\begin{array}{l} = \int {\sqrt {{{\left( {x + 1} \right)}^2} + 2} \,dx} \, - \dfrac{1}{2}\int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} \,\,dx\,\, - \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + 2} }}} \,\,dx\\ \\ \,\,\,\,\left( \begin{array}{l} {\rm{of \,the \,standard\, }}\\ {\rm{form}}\,\left( {26} \right) \end{array} \right) \,\,\,\,\,\, \left( \begin{array}{l} {\rm{use the substituion}}\\ {x^2} + 2x + 3 = t \end{array} \right) \,\,\,\,\, \left( \begin{array}{l} {\rm{of\, the\, standard\, }}\\ {\rm{\,form}}\,\left( {22} \right) \end{array} \right) \end{array}$
	$= \dfrac{1}{2}\left\{ {\left( {x + 1} \right)\sqrt {{x^2} + 2x + 3} + 2\ln \left\| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right\|} \right\} - \sqrt {{x^2} + 2x + 3} - \ln \left\{ {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} } \right\} + C$
	$= \dfrac{1}{2}\left( {x + 1} \right)\sqrt {{x^2} + 2x + 3} \,\,\, - \,\,\,\sqrt {{x^2} + 2x + 3} \,\,\,\, + \,\,\,C$
	$= \dfrac{1}{2}\left( {x - 1} \right)\sqrt {{x^2} + 2x + 3} \,\, + \,\,\,C$

Solution: 1-(b)

Using the same values for $\alpha ,\beta \,$ and $\gamma$ this integral now becomes

	${I_2} = {\int {\left( {{x^2} + 2x + 3} \right)} ^{3/2}}dx - \dfrac{1}{2}\int {\left( {2x + 2} \right)\sqrt {\underbrace{{x^2} + 2x + 3}_{\rm{Comment:1}}} } \,\,dx - \int {\sqrt {\underbrace{{x^2} + 2x + 3}_{\rm{Comment:2}}} } \,\,dx$
	Comment:1 Use the substitution ${x^2} + 2x + 3=t$
	Comment:2 (of the standard form $(26)$

The last two integrals can be evaluated as indicated.

To evaluate ${I_3} = {\int {\left( {{x^2} + 2x + 3} \right)} ^{3/2}}dx = \int {{{\left( {{{\left( {x + 1} \right)}^2} + 2} \right)}^{3/2}}} dx,$

we let $x + 1 = \sqrt 2 \tan \theta$ so that $dx = \sqrt 2 {\sec ^2}\theta d\theta$ :

	$\Rightarrow\,\,\,\, {I_3} = \int {{2^{3/2}}} {\sec ^3}\theta .\sqrt 2 {\sec ^2}\theta d\theta :$
	$= 4\int {{{\sec }^5}\theta d\theta }$
	$I = \dfrac{{{I_3}}}{4} = \int {\mathop {{{\sec }^3}\theta }\limits_{\rm{Ist}} .\mathop {{{\sec }^2}}\limits_{\rm{IInd}} d\theta }$
	$= {\sec ^3}\theta \tan \theta - 3\int {{{\sec }^3}\theta } {\tan ^2}\theta d\theta$
	$= {\sec ^3}\theta \tan \theta - 3\int {{{\sec }^3}\theta } \left( {{{\sec }^2}\theta - 1} \right)d\theta$
	$= {\sec ^3}\theta \tan \theta - 3\int {{{\sec }^5}\theta d\theta + 3} \int {{{\sec }^3}\theta d\theta }$
	$= {\sec ^3}\theta \tan \theta - 3I + 3{I_4}$

where ${I_4} = \int {{{\sec }^3}\theta d\theta }$

$\Rightarrow \,\,\,\,4I = {\sec ^3}\theta \tan \theta + 3{I_4}$

To evaluate ${I_4}$ , we again use integration by parts:

	${I_4} = \int {\mathop {\sec \theta }\limits_{\rm{Ist}} \mathop {{{\sec }^2}\theta }\limits_{\rm{IInd} }} d\theta$
	$= \sec \theta \tan \theta - \int {\sec \theta {{\tan }^2}\theta d\theta }$
	$= \sec \theta \tan \theta - \int {\sec \theta \left( {{{\sec }^2}\theta - 1} \right)d\theta }$
	$= \sec \theta \tan \theta - {I_4}\,\,\,\,\,\, + \,\,\,\int {\sec \theta d\theta }$
	$\Rightarrow \,\,\,\,2{I_4} = \sec \theta \tan \theta + \ln \left\| {\sec \theta + \tan \theta } \right\|$
	$\Rightarrow\,\,\,\, I = \dfrac{1}{4}{\sec ^3}\theta \tan \theta + \dfrac{3}{8}\sec \theta \tan \theta + \dfrac{3}{8}\ln \left\| {\sec \theta + \tan \theta } \right\| + C$

Using $I$ , ${I_2}$ can be determined.

Monday, 4 August 2014

CHAPTER 15- Worked Out Examples

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