Solution 2.2:4a - Förberedande kurs i matematik 2

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                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 2.2:4a
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_4a.gif </center> {{NAVCONTENT_STOP}})
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- {{NAVCONTENT_START}} + What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,
- <center> [[Bild:2_2_4a.gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}
  +  
  + we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
  + &= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
  + = \left\{\begin{align}
  + u &= x/2\\[5pt]
  + du &= \tfrac{1}{2}\,dx
  + \end{align}\right\}\\[5pt] 
  + &= \frac{1}{4}\int \frac{2\,du}{u^2+1}
  + = \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt] 
  + &= \frac{1}{2}\arctan u + C
  + = \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.} 
  + \end{align}</math>}}
Current revision
What makes our integral differ from that in the exercise´s text is that there is a term \displaystyle x^2+4 instead of \displaystyle x^2+1, but if we factor out the 4 from the denominator,





\displaystyle \int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}


we obtain the correct second term in the denominator. On the other hand, there is no longer \displaystyle x^2 but \displaystyle \tfrac{1}{4}x^2, although we can get around this by substituting \displaystyle u=\tfrac{1}{2}x,





\displaystyle \begin{align}
\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
= \left\{\begin{align}
u &= x/2\\[5pt]
du &= \tfrac{1}{2}\,dx
\end{align}\right\}\\[5pt] 
&= \frac{1}{4}\int \frac{2\,du}{u^2+1}
= \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt] 
&= \frac{1}{2}\arctan u + C
= \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.} 
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_2.2:4a"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,
-<center> [[Bild:2_2_4a.gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}
+we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,
+{{Displayed math||<math>\begin{align}
+\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
+&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
+= \left\{\begin{align}
+u &= x/2\\[5pt]
+du &= \tfrac{1}{2}\,dx
+\end{align}\right\}\\[5pt]
+&= \frac{1}{4}\int \frac{2\,du}{u^2+1}
+= \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt]
+&= \frac{1}{2}\arctan u + C
+= \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}
 \displaystyle \begin{align}
\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
= \left\{\begin{align}
u &= x/2\\[5pt]
du &= \tfrac{1}{2}\,dx
\end{align}\right\}\\[5pt] 
&= \frac{1}{4}\int \frac{2\,du}{u^2+1}
= \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt] 
&= \frac{1}{2}\arctan u + C
= \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.} 
\end{align}