Lösung 2.3:2c

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If we use the definition of
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<center> [[Image:2_3_2c.gif]] </center>
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<math>\tan x</math>
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and write the integral as
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<math>\int{\tan x\,dx}=\int{\frac{\sin x}{\cos x}\,dx}</math>
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we see that the numerator
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<math>\sin x</math>
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is the derivative of the denominator (apart from the minus sign). Hence, the substitution
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<math>u=\cos x</math>
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will work,
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<math>\begin{align}
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& \int{\frac{\sin x}{\cos x}\,dx}=\left\{ \begin{matrix}
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u=\cos x \\
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du=\left( \cos x \right)^{\prime }\,dx=-\sin x\,dx \\
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\end{matrix} \right\} \\
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& =-\int{\frac{\,du}{u}}=-\ln \left| u \right|+C=-\ln \left| \cos x \right|+C \\
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\end{align}</math>
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NOTE:
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<math>-\ln \left| \cos x \right|+C</math>
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is only a primitive function in intervals in which
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<math>\cos x\ne 0</math>.

Version vom 14:16, 22. Okt. 2008

If we use the definition of \displaystyle \tan x and write the integral as


\displaystyle \int{\tan x\,dx}=\int{\frac{\sin x}{\cos x}\,dx}


we see that the numerator \displaystyle \sin x is the derivative of the denominator (apart from the minus sign). Hence, the substitution \displaystyle u=\cos x will work,


\displaystyle \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}


NOTE: \displaystyle -\ln \left| \cos x \right|+C is only a primitive function in intervals in which \displaystyle \cos x\ne 0.

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