From Förberedande kurs i matematik 2

Revision as of 14:48, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:2c moved to Solution 2.3:2c: Robot: moved page) ← Previous diff		Current revision (08:59, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we use the definition of <math>\tan x</math> and write the integral as
-	<center> [[Image:2_3_2c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed 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>}}
		+
		+
		+	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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Current revision

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.