Lösung 2.2:3a

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The secret behind a successful substitution is to be able to recognize the integral as an expression of the type
The secret behind a successful substitution is to be able to recognize the integral as an expression of the type
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{{Displayed math||<math>\int \left( \begin{matrix}
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\text{an expression}\\
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\text{in u}
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\end{matrix}\right)\cdot u'\,dx\,,</math>}}
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<math>\int{\left( \begin{matrix}
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where <math>u=u(x)</math> is the actual substitution. In the integral
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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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{{Displayed math||<math>\int 2x\sin x^2\,dx</math>}}
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<math>u=u\left( x \right)</math>
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is the actual substitution. In the integral
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 +
we see that the expression <math>x^2</math> is the argument for the sine function, as the same time as its derivative <math>\bigl(x^2\bigr)'=2x</math> 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
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<math>\int{2x\sin x^{2}\,dx}</math>
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{{Displayed math||<math>\int u'\sin u\,dx\,\textrm{.}</math>}}
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Thus, we can use <math>u=x^2</math> for the substitution,
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we see that the expression
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{{Displayed math||<math>\begin{align}
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<math>x^{2}</math>
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\int 2x\sin x^2\,dx
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is the argument for the sine function, as the same time as its derivative
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&=\left\{\begin{align}
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<math>\left( x^{2} \right)^{\prime }=2x</math>
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u &= x^2\\[5pt]
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stands as a factor in front of sine. Therefore, if we set
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du &= 2x\,dx
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<math>u=x^{2}</math>, the integral, the integral will be of the form
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\end{align}\right\} = \int{\sin u\,du}\\[5pt]
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&= -\cos u+C = -\cos x^2 + C\,\textrm{.}
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\end{align}</math>}}
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<math>\int{{u}'\sin u\,dx}</math>
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Thus, we can use
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<math>u=x^{2}</math>
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for the substitution:
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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>
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Version vom 14:04, 28. Okt. 2008

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

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

\text{an expression}\\ \text{in u} \end{matrix}\right)\cdot u'\,dx\,,

where \displaystyle u=u(x) is the actual substitution. In the integral

\displaystyle \int 2x\sin x^2\,dx

we see that the expression \displaystyle x^2 is the argument for the sine function, as the same time as its derivative \displaystyle \bigl(x^2\bigr)'=2x stands as a factor in front of sine. Therefore, if we set \displaystyle u=x^2, the integral, the integral will be of the form

\displaystyle \int u'\sin u\,dx\,\textrm{.}

Thus, we can use \displaystyle u=x^2 for the substitution,

\displaystyle \begin{align}

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

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