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Lösung 2.2:3f

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
Let's rewrite the integral somewhat,
Let's rewrite the integral somewhat,
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{{Displayed math||<math>2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>2\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
Here, we see that the factor on the right, <math>1/2\sqrt{x}</math>, is the derivative of the expression <math>\sqrt{x}</math>, which appears in the factor on the left, <math>2\sin \sqrt{x}\,</math>. With the substitution <math>u=\sqrt{x}</math>, the integrand can therefore be written as
Here, we see that the factor on the right, <math>1/2\sqrt{x}</math>, is the derivative of the expression <math>\sqrt{x}</math>, which appears in the factor on the left, <math>2\sin \sqrt{x}\,</math>. With the substitution <math>u=\sqrt{x}</math>, the integrand can therefore be written as
-
{{Displayed math||<math>2\sin u\cdot u'</math>}}
+
{{Abgesetzte Formel||<math>2\sin u\cdot u'</math>}}
and the integral becomes
and the integral becomes
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx
\int \frac{\sin \sqrt{x}}{\sqrt{x}}\,dx
&= \left\{ \begin{align}
&= \left\{ \begin{align}

Version vom 13:02, 10. Mär. 2009

Let's rewrite the integral somewhat,

2sinx12x.

Here, we see that the factor on the right, 12x , is the derivative of the expression x , which appears in the factor on the left, 2sinx . With the substitution u=x , the integrand can therefore be written as

2sinuu

and the integral becomes

xsinxdx=udu=x=(x)dx=12xdx=2sinudu=−2cosu+C=−2cosx+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3f“