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Lösung 2.3:2c

Aus Online Mathematik Brückenkurs 2

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Version vom 10:27, 11. Mär. 2009

If we use the definition of tanx and write the integral as

tanxdx=sinxcosxdx 

we see that the numerator sinx is the derivative of the denominator (apart from the minus sign). Hence, the substitution u=cosx will work,

sinxcosxdx=udu=cosx=(cosx)dx=−sinxdx=−udu=−lnu+C=−lncosx+C.


Note: −lncosx+C is only a primitive function in intervals in which cosx=0.

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