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

Lösung 2.3:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
If we use the definition of
+
If we use the definition of <math>\tan x</math> and write the integral as
-
<math>\tan x</math>
+
-
and write the integral as
+
 +
{{Displayed math||<math>\int\tan x\,dx = \int\frac{\sin x}{\cos x}\,dx</math>}}
-
<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,
 +
{{Displayed math||<math>\begin{align}
 +
\int\frac{\sin x}{\cos x}\,dx
 +
&= \left\{\begin{align}
 +
u &= \cos x\\[5pt]
 +
du &= (\cos x)'\,dx = -\sin x\,dx
 +
\end{align}\right\}\\[5pt]
 +
&= -\int\frac{du}{u}\\[5pt]
 +
&= -\ln |u| + C\\[5pt]
 +
&= -\ln |\cos x| + C\,\textrm{.}
 +
\end{align}</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,
 
-
 
+
Note: <math>-\ln \left| \cos x \right|+C</math> is only a primitive function in intervals in which <math>\cos x\ne 0</math>.
-
<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>.
+

Version vom 08:59, 29. 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=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“