Lösung 2.3:1b
Aus Online Mathematik Brückenkurs 2
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Version vom 10:26, 11. 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} |
