Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 2.3:1c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
K (Solution 2.3:1c moved to Lösung 2.3:1c: Robot: moved page)

Version vom 10:26, 11. 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“