Lösung 2.3:1c

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The integrand consists of two factors, so partial integration is a plausible method. The most obvious thing to do is to choose
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The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose <math>x^2</math> as the factor that we will differentiate and <math>\cos x</math> as the factor that we will integrate. Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,
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<math>x^{2}</math>
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as the factor that we will differentiate and
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<math>\cos x</math>
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as the factor that we will integrate. Admittedly, the
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<math>x^{2}</math>
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-factor will not be differentiated away, but its exponent decreases by
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<math>\text{1}</math>
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and this makes the integral a little easier:
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{{Displayed math||<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}
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<math>\int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\int{2x\centerdot \sin x\,dx}}</math>
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We can attack the integral on the right-hand side in the same way. Let <math>2x</math> be the factor that we differentiate and <math>\sin x</math> the factor that we integrate. This time, we have only one factor left,
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We can attack the integral on the right-hand side in the same way. Let
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<math>2x</math>
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be the factor that we differentiate and
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<math>\sin x</math>
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the factor that we integrate. This time, we have only one factor left:
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<math>\begin{align}
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& \int{2x\centerdot \sin x\,dx}=2x\centerdot \left( -\cos x \right)-\int{2\centerdot }\left( -\cos x \right)\,dx \\
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& =-2x\cos x+2\int{\cos x\,dx} \\
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& =-2x\cos x+2\sin x+C \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\int 2x\cdot \sin x\,dx
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&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]
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&= -2x\cos x + 2\int \cos x\,dx\\[5pt]
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&= -2x\cos x + 2\sin x + C\,\textrm{.}
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\end{align}</math>}}
All in all, we obtain
All in all, we obtain
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\int x^2\cos x\,dx
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& \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\left( -2x\cos x+2\sin x+C \right)} \\
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&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt]
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& =x^{2}\centerdot \sin x+2x\cos x-2\sin x+C \\
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&= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.
For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

Version vom 08:22, 29. Okt. 2008

The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose \displaystyle x^2 as the factor that we will differentiate and \displaystyle \cos x as the factor that we will integrate. Admittedly, the \displaystyle x^2-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,

\displaystyle \int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}

We can attack the integral on the right-hand side in the same way. Let \displaystyle 2x be the factor that we differentiate and \displaystyle \sin x the factor that we integrate. This time, we have only one factor left,

\displaystyle \begin{align}

\int 2x\cdot \sin x\,dx &= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt] &= -2x\cos x + 2\int \cos x\,dx\\[5pt] &= -2x\cos x + 2\sin x + C\,\textrm{.} \end{align}

All in all, we obtain

\displaystyle \begin{align}

\int x^2\cos x\,dx &= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt] &= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.} \end{align}

For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

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