Lösung 2.3:1c
Aus Online Mathematik Brückenkurs 2
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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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}} |
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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int 2x\cdot \sin x\,dx | \int 2x\cdot \sin x\,dx | ||
&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt] | &= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt] | ||
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All in all, we obtain | All in all, we obtain | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int x^2\cos x\,dx | \int x^2\cos x\,dx | ||
&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt] | &= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt] | ||
Version vom 13:03, 10. Mär. 2009
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.
