Lösung 2.3:1b

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we look at the formula for partial integration,
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If we look at the formula for integration by parts,
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{{Displayed math||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
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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 <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> 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 <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!
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{{Displayed math||<math>\begin{align}
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we see that if we choose
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\int (x+1)\sin x\,dx
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<math>f\left( x \right)=\text{sin }x\text{ }</math>
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&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]
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and
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&= -(x+1)\cos x + \int \cos x\,dx\\[5pt]
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<math>g\left( x \right)=x+\text{1}</math>, then the factor
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&= -(x+1)\cos x + \sin x + C\,\textrm{.}
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<math>g\left( x \right)</math>
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\end{align}</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>
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Version vom 08:13, 29. Okt. 2008

If we look at the formula for integration by parts,

\displaystyle \int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,

we see that if we choose \displaystyle f(x)=\sin x and \displaystyle g(x)=x+1, then the factor \displaystyle g(x) 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(x) (which we can) and that we can then integrate it. Let's try!

\displaystyle \begin{align}

\int (x+1)\sin x\,dx &= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt] &= -(x+1)\cos x + \int \cos x\,dx\\[5pt] &= -(x+1)\cos x + \sin x + C\,\textrm{.} \end{align}

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