Lösung 2.2:3b

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If we to succeed in simplifying the integral with a substitution, we must find an expression
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<center> [[Image:2_2_3b.gif]] </center>
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<math>u=u\left( x \right)</math>
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so that the integral can be written as
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<math>\int{\left( \begin{matrix}
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\text{something} \\
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\text{in}\quad u \\
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\end{matrix} \right)}\centerdot {u}'\,dx</math>.
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As our integral is written,
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<math>\int{\sin x\cos x\,dx}</math>
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we see that the second factor
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<math>\cos x</math>
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is a derivative of the first factor,
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<math>\sin x</math>. If
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<math>u=\text{sin }x</math>, the integral can thus be written as
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<math>\int{u\centerdot {u}'\,dx}</math>
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and this makes
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<math>u=\text{sin }x</math>
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an appropriate substitution,
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<math>\begin{align}
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& \int{\sin x\cos x\,dx}=\left\{ \begin{matrix}
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u=\text{sin }x \\
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du=\left( \sin x \right)^{\prime }\,dx=\cos x\,dx \\
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\end{matrix} \right\} \\
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& =\int{u\,du=\frac{1}{2}u^{2}} \\
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& =\frac{1}{2}\sin ^{2}x+C \\
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\end{align}</math>

Version vom 10:28, 21. Okt. 2008

If we to succeed in simplifying the integral with a substitution, we must find an expression \displaystyle u=u\left( x \right) so that the integral can be written as


\displaystyle \int{\left( \begin{matrix} \text{something} \\ \text{in}\quad u \\ \end{matrix} \right)}\centerdot {u}'\,dx.

As our integral is written,


\displaystyle \int{\sin x\cos x\,dx}


we see that the second factor \displaystyle \cos x is a derivative of the first factor, \displaystyle \sin x. If \displaystyle u=\text{sin }x, the integral can thus be written as


\displaystyle \int{u\centerdot {u}'\,dx}

and this makes \displaystyle u=\text{sin }x an appropriate substitution,


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

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