Lösung 2.2:3f

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K (Lösning 2.2:3f moved to Solution 2.2:3f: Robot: moved page)
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{{NAVCONTENT_START}}
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Let's rewrite the integral somewhat:
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<center> [[Image:2_2_3f.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>2\sin \sqrt{x}\centerdot \frac{1}{2\sqrt{x}}</math>
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Here, we see that the factor on the right,
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<math>\frac{1}{2\sqrt{x}}</math>
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is the derivative of the expression
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<math>\sqrt{x}</math>, which appears in the factor on the left,
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<math>2\sin \sqrt{x}</math>
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With the substitution
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<math>u=\sqrt{x}</math>, the integrand can therefore be written as
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<math>2\sin u\centerdot {u}'</math>
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and the integral becomes
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<math>\begin{align}
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& \int{\frac{\sin \sqrt{x}}{\sqrt{x}}}\,dx=\left\{ \begin{matrix}
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u=\sqrt{x} \\
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du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\
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\end{matrix}\, \right\} \\
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& =2\int{\sin u\,du} \\
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& =-2\cos u+C \\
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& =-2\cos \sqrt{x}+C \\
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\end{align}</math>

Version vom 15:38, 28. Okt. 2008

Let's rewrite the integral somewhat:


\displaystyle 2\sin \sqrt{x}\centerdot \frac{1}{2\sqrt{x}}


Here, we see that the factor on the right, \displaystyle \frac{1}{2\sqrt{x}} is the derivative of the expression \displaystyle \sqrt{x}, which appears in the factor on the left, \displaystyle 2\sin \sqrt{x} With the substitution \displaystyle u=\sqrt{x}, the integrand can therefore be written as


\displaystyle 2\sin u\centerdot {u}'


and the integral becomes


\displaystyle \begin{align} & \int{\frac{\sin \sqrt{x}}{\sqrt{x}}}\,dx=\left\{ \begin{matrix} u=\sqrt{x} \\ du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\ \end{matrix}\, \right\} \\ & =2\int{\sin u\,du} \\ & =-2\cos u+C \\ & =-2\cos \sqrt{x}+C \\ \end{align}

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