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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.
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