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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If we are to succeed in simplifying the integral with a substitution, we must find an expression <math>u = u(x)</math> so that the integral can be written as
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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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{{Displayed math||<math>\int \left(\begin{matrix}
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<math>\int{\left( \begin{matrix}
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\text{something}\\
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\text{something} \\
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\text{in u}
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\text{in}\quad u \\
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\end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}</math>}}
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\end{matrix} \right)}\centerdot {u}'\,dx</math>.
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As our integral is written,
As our integral is written,
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{{Displayed math||<math>\int\sin x\cos x\,dx</math>}}
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<math>\int{\sin x\cos x\,dx}</math>
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we see that the second factor <math>\cos x</math> is a derivative of the first factor, <math>\sin x</math>. If <math>u=\sin x</math>, the integral can thus be written as
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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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{{Displayed math||<math>\int u\cdot u'\,dx</math>}}
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<math>u=\text{sin }x</math>
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an appropriate substitution,
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and this makes <math>u=\sin x</math> an appropriate substitution,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \int{\sin x\cos x\,dx}=\left\{ \begin{matrix}
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\int \sin x\cos x\,dx
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u=\text{sin }x \\
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&= \left\{ \begin{align}
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du=\left( \sin x \right)^{\prime }\,dx=\cos x\,dx \\
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u &= \sin x\\[5pt]
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\end{matrix} \right\} \\
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du &= (\sin x)'\,dx = \cos x\,dx
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& =\int{u\,du=\frac{1}{2}u^{2}} \\
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\end{align} \right\}\\[5pt]
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& =\frac{1}{2}\sin ^{2}x+C \\
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&= \int u\,du\\[5pt]
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\end{align}</math>
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&= \frac{1}{2}u^{2} + C\\[5pt]
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&= \frac{1}{2}\sin^2\!x + C\,\textrm{.}
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\end{align}</math>}}

Version vom 14:15, 28. Okt. 2008

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

\displaystyle \int \left(\begin{matrix}

\text{something}\\ \text{in u} \end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}

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=\sin x, the integral can thus be written as

\displaystyle \int u\cdot u'\,dx

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

\displaystyle \begin{align}

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

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