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Lösung 2.2:3a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 2.2:3a moved to Solution 2.2:3a: Robot: moved page)
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{{NAVCONTENT_START}}
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The secret behind a successful substitution is to be able to recognize the integral as an expression of the type
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<center> [[Image:2_2_3a.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\int{\left( \begin{matrix}
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\text{an}\quad \text{expression} \\
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\text{in}\quad u \\
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\end{matrix} \right)}\centerdot {u}'\,dx</math>,
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where
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<math>u=u\left( x \right)</math>
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is the actual substitution. In the integral
 +
 
 +
 
 +
<math>\int{2x\sin x^{2}\,dx}</math>
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 +
 
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we see that the expression
 +
<math>x^{2}</math>
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is the argument for the sine function, as the same time as its derivative
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<math>\left( x^{2} \right)^{\prime }=2x</math>
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stands as a factor in front of sine. Therefore, if we set
 +
<math>u=x^{2}</math>, the integral, the integral will be of the form
 +
 
 +
 
 +
<math>\int{{u}'\sin u\,dx}</math>
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 +
 
 +
Thus, we can use
 +
<math>u=x^{2}</math>
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for the substitution:
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 +
 
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<math>\begin{align}
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& \int{2x\sin x^{2}\,dx}=\left\{ \begin{matrix}
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u=x^{2} \\
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du=2x\,dx \\
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\end{matrix} \right\}=\int{\sin u\,du} \\
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& =-\cos u+C=-\cos x^{2}+C \\
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\end{align}</math>

Version vom 12:51, 20. Okt. 2008

The secret behind a successful substitution is to be able to recognize the integral as an expression of the type


anexpressioninuudx ,

where u=ux  is the actual substitution. In the integral


2xsinx2dx 


we see that the expression x2 is the argument for the sine function, as the same time as its derivative x2=2x  stands as a factor in front of sine. Therefore, if we set u=x2, the integral, the integral will be of the form


usinudx 


Thus, we can use u=x2 for the substitution:


2xsinx2dx=u=x2du=2xdx=sinudu=−cosu+C=−cosx2+C 

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3a“