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Lösung 2.2:3f

Aus Online Mathematik Brückenkurs 2

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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=u=xdu=xdx=12xdx=2sinudu=−2cosu+C=−2cosx+C

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3f“