Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 2.3:1c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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,
-
{{Displayed math||<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}
+
{{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,
-
{{Displayed math||<math>\begin{align}
+
{{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]
Zeile 14: Zeile 14:
All in all, we obtain
All in all, we obtain
-
{{Displayed math||<math>\begin{align}
+
{{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 x2 as the factor that we will differentiate and cosx as the factor that we will integrate. Admittedly, the x2-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,

x2cosxdx=x2sinx−2xsinxdx. 

We can attack the integral on the right-hand side in the same way. Let 2x be the factor that we differentiate and sinx the factor that we integrate. This time, we have only one factor left,

2xsinxdx=2x(−cosx)−2(−cosx)dx=−2xcosx+2cosxdx=−2xcosx+2sinx+C.

All in all, we obtain

x2cosxdx=x2sinx−(−2xcosx+2sinx+C)=x2sinx+2xcosx−2sinx+C. 

For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1c“