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Lösung 2.3:1c

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.3:1c moved to Solution 2.3:1c: Robot: moved page)
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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 x2 as the factor that we will differentiate and cosx as the factor that we will integrate. Admittedly, the x2 -factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier:


x2cosxdx=x2sinx−2xsinxdx 


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


2xsinxdx=2x−cosx−2−cosxdx=−2xcosx+2cosxdx=−2xcosx+2sinx+C


All in all, we obtain


x2cosxdx=x2sinx−−2xcosx+2sinx+C=x2sinx+2xcosx−2sinx+C 


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

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