Lösung 2.2:3f

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Let's rewrite the integral somewhat:
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Let's rewrite the integral somewhat,
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{{Displayed math||<math>2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
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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, <math>1/2\sqrt{x}</math>, is the derivative of the expression <math>\sqrt{x}</math>, which appears in the factor on the left, <math>2\sin \sqrt{x}\,</math>. With the substitution <math>u=\sqrt{x}</math>, the integrand can therefore be written as
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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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{{Displayed math||<math>2\sin u\cdot u'</math>}}
and the integral becomes
and the integral becomes
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx
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& \int{\frac{\sin \sqrt{x}}{\sqrt{x}}}\,dx=\left\{ \begin{matrix}
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&= \left\{ \begin{align}
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u=\sqrt{x} \\
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u &= \sqrt{x}\\[5pt]
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du=\left( \sqrt{x} \right)^{\prime }\,dx=\frac{1}{2\sqrt{x}}\,dx \\
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du &= (\sqrt{x}\,)'\,dx = \frac{1}{2\sqrt{x}}\,dx
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\end{matrix}\, \right\} \\
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\end{align}\, \right\}\\[5pt]
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& =2\int{\sin u\,du} \\
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&= 2\int \sin u\,du\\[5pt]
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& =-2\cos u+C \\
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&= -2\cos u+C\\[5pt]
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& =-2\cos \sqrt{x}+C \\
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&= -2\cos\sqrt{x} + C\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 15:43, 28. Okt. 2008

Let's rewrite the integral somewhat,

\displaystyle 2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}

Here, we see that the factor on the right, \displaystyle 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\cdot u'

and the integral becomes

\displaystyle \begin{align}

\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx &= \left\{ \begin{align} u &= \sqrt{x}\\[5pt] du &= (\sqrt{x}\,)'\,dx = \frac{1}{2\sqrt{x}}\,dx \end{align}\, \right\}\\[5pt] &= 2\int \sin u\,du\\[5pt] &= -2\cos u+C\\[5pt] &= -2\cos\sqrt{x} + C\,\textrm{.} \end{align}

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