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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<center> [[Image:2_3_1c-1(2).gif]] </center>
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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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<center> [[Image:2_3_1c-2(2).gif]] </center>
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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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<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
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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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All in all, we obtain
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<math>\begin{align}
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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}\centerdot \sin x+2x\cos x-2\sin x+C \\
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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.

Version vom 13:40, 21. Okt. 2008

The integrand consists of two factors, so partial integration 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 \displaystyle \text{1} and this makes the integral a little easier:


\displaystyle \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\int{2x\centerdot \sin x\,dx}}


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\centerdot \sin x\,dx}=2x\centerdot \left( -\cos x \right)-\int{2\centerdot }\left( -\cos x \right)\,dx \\ & =-2x\cos x+2\int{\cos x\,dx} \\ & =-2x\cos x+2\sin x+C \\ \end{align}


All in all, we obtain


\displaystyle \begin{align} & \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\left( -2x\cos x+2\sin x+C \right)} \\ & =x^{2}\centerdot \sin x+2x\cos x-2\sin x+C \\ \end{align}


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

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