Lösung 2.3:2c
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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If we use the definition of <math>\tan x</math> and write the integral as | If we use the definition of <math>\tan x</math> and write the integral as | ||
| - | {{ | + | {{Abgesetzte Formel||<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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int\frac{\sin x}{\cos x}\,dx | \int\frac{\sin x}{\cos x}\,dx | ||
&= \left\{\begin{align} | &= \left\{\begin{align} | ||
Version vom 13:04, 10. Mär. 2009
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{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} |
Note: \displaystyle -\ln \left| \cos x \right|+C is only a primitive function in intervals in which \displaystyle \cos x\ne 0.
