Lösung 2.3:1b

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we look at the formula for partial integration,
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<center> [[Image:2_3_1b.gif]] </center>
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<math>\int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx}</math>
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we see that if we choose
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<math>f\left( x \right)=\text{sin }x\text{ }</math>
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and
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<math>g\left( x \right)=x+\text{1}</math>, then the factor
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<math>g\left( x \right)</math>
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will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for
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<math>f\left( x \right)</math>
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(which we can) and that we can then integrate it. Let's try!
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<math>\begin{align}
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& \int{\left( x+1 \right)}\sin x\,dx=\left( x+1 \right)\centerdot \left( -\cos x \right)-\int{1\centerdot }\left( -\cos x \right)\,dx \\
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& =-\left( x+1 \right)\cos x+\int{\cos x}\,dx \\
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& =-\left( x+1 \right)\cos x+\sin x+C \\
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\end{align}</math>

Version vom 13:32, 21. Okt. 2008

we look at the formula for partial integration,


\displaystyle \int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx}


we see that if we choose \displaystyle f\left( x \right)=\text{sin }x\text{ } and \displaystyle g\left( x \right)=x+\text{1}, then the factor \displaystyle g\left( x \right) will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for \displaystyle f\left( x \right) (which we can) and that we can then integrate it. Let's try!


\displaystyle \begin{align} & \int{\left( x+1 \right)}\sin x\,dx=\left( x+1 \right)\centerdot \left( -\cos x \right)-\int{1\centerdot }\left( -\cos x \right)\,dx \\ & =-\left( x+1 \right)\cos x+\int{\cos x}\,dx \\ & =-\left( x+1 \right)\cos x+\sin x+C \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1b“