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Aus Online Mathematik Brückenkurs 2

Version vom 14:48, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.3:2c moved to Solution 2.3:2c: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 14:16, 22. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
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-	{{NAVCONTENT_START}}	+	If we use the definition of
-	<center> [[Image:2_3_2c.gif]] </center>	+	<math>\tan x</math>
-	{{NAVCONTENT_STOP}}	+	and write the integral as
		+
		+
		+	<math>\int{\tan x\,dx}=\int{\frac{\sin x}{\cos x}\,dx}</math>
		+
		+
		+	we see that the numerator
		+	<math>\sin x</math>
		+	is the derivative of the denominator (apart from the minus sign). Hence, the substitution
		+	<math>u=\cos x</math>
		+	will work,
		+
		+
		+	<math>\begin{align}
		+	& \int{\frac{\sin x}{\cos x}\,dx}=\left\{ \begin{matrix}
		+	u=\cos x \\
		+	du=\left( \cos x \right)^{\prime }\,dx=-\sin x\,dx \\
		+	\end{matrix} \right\} \\
		+	& =-\int{\frac{\,du}{u}}=-\ln \left| u \right|+C=-\ln \left| \cos x \right|+C \\
		+	\end{align}</math>
		+
		+
		+	NOTE:
		+	<math>-\ln \left| \cos x \right|+C</math>
		+	is only a primitive function in intervals in which
		+	<math>\cos x\ne 0</math>.

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Version vom 14:16, 22. Okt. 2008

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=u=cosxdu=cosxdx=−sinxdx=−udu=−lnu+C=−lncosx+C 

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