Lösung 2.2:3f
Aus Online Mathematik Brückenkurs 2
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K (Solution 2.2:3f moved to Lösung 2.2:3f: Robot: moved page) |
Version vom 10:24, 11. Mär. 2009
Let's rewrite the integral somewhat,
| \displaystyle 2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.} |
Here, we see that the factor on the right, \displaystyle 1/2\sqrt{x}, is the derivative of the expression \displaystyle \sqrt{x}, which appears in the factor on the left, \displaystyle 2\sin \sqrt{x}\,. With the substitution \displaystyle u=\sqrt{x}, the integrand can therefore be written as
| \displaystyle 2\sin u\cdot u' |
and the integral becomes
| \displaystyle \begin{align}
\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx &= \left\{ \begin{align} u &= \sqrt{x}\\[5pt] du &= (\sqrt{x}\,)'\,dx = \frac{1}{2\sqrt{x}}\,dx \end{align}\, \right\}\\[5pt] &= 2\int \sin u\,du\\[5pt] &= -2\cos u+C\\[5pt] &= -2\cos\sqrt{x} + C\,\textrm{.} \end{align} |
