Lösung 2.3:1b
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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If we look at the formula for integration by parts, | If we look at the formula for integration by parts, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}} |
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! | 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! | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int (x+1)\sin x\,dx | \int (x+1)\sin x\,dx | ||
&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt] | &= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt] | ||
Version vom 13:03, 10. Mär. 2009
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} |
