Solution 2.2:4c - 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 2.2:4c
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_4c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:2_2_4c-2(2).gif </center> {{NAVCONTENT_STOP}})
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- {{NAVCONTENT_START}} + The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
- <center> [[Bild:2_2_4c-1(2).gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}
- {{NAVCONTENT_START}} +  
- <center> [[Bild:2_2_4c-2(2).gif]] </center> + We take out a factor 4 from the denominator
- {{NAVCONTENT_STOP}} +  
  + {{Displayed math||<math>\int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}</math>}}
  +  
  + and rewrite the quadratic term as
  +  
  + {{Displayed math||<math>\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}</math>}}
  +  
  + If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
  + &= \left\{\begin{align}
  + u &= (x+2)/2\\[5pt]
  + du &= dx/2
  + \end{align}\right\}\\[5pt] 
  + &= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt]
  + &= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt] 
  + &= \frac{1}{2}\arctan u + C\\[5pt] 
  + &= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.}
  + \end{align}</math>}}
Current revision
The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,





\displaystyle \int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}


We take out a factor 4 from the denominator





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


and rewrite the quadratic term as





\displaystyle \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}


If we now substitute \displaystyle u = (x+2)/2, we obtain the integral in the exercise





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




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_2.2:4c"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
-<center> [[Bild:2_2_4c-1(2).gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}
-{{NAVCONTENT_START}}
-<center> [[Bild:2_2_4c-2(2).gif]] </center>
+We take out a factor 4 from the denominator
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}</math>}}
+and rewrite the quadratic term as
+{{Displayed math||<math>\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}</math>}}
+If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
+{{Displayed math||<math>\begin{align}
+\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
+&= \left\{\begin{align}
+u &= (x+2)/2\\[5pt]
+du &= dx/2
+\end{align}\right\}\\[5pt]
+&= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt]
+&= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt]
+&= \frac{1}{2}\arctan u + C\\[5pt]
+&= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}
 \displaystyle \int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}
 \displaystyle \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}
 \displaystyle \begin{align}
\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
&= \left\{\begin{align}
u &= (x+2)/2\\[5pt]
du &= dx/2
\end{align}\right\}\\[5pt] 
&= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt]
&= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt] 
&= \frac{1}{2}\arctan u + C\\[5pt] 
&= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.}
\end{align}