From Förberedande kurs i matematik 2

Revision as of 14:43, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:3b moved to Solution 2.2:3b: Robot: moved page) ← Previous diff		Current revision (14:15, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we are to succeed in simplifying the integral with a substitution, we must find an expression <math>u = u(x)</math> so that the integral can be written as
-	<center> [[Image:2_2_3b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\int \left(\begin{matrix}
		+	\text{something}\\
		+	\text{in u}
		+	\end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}</math>}}
		+
		+	As our integral is written,
		+
		+	{{Displayed math||<math>\int\sin x\cos x\,dx</math>}}
		+
		+	we see that the second factor <math>\cos x</math> is a derivative of the first factor, <math>\sin x</math>. If <math>u=\sin x</math>, the integral can thus be written as
		+
		+	{{Displayed math||<math>\int u\cdot u'\,dx</math>}}
		+
		+	and this makes <math>u=\sin x</math> an appropriate substitution,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int \sin x\cos x\,dx
		+	&= \left\{ \begin{align}
		+	u &= \sin x\\[5pt]
		+	du &= (\sin x)'\,dx = \cos x\,dx
		+	\end{align} \right\}\\[5pt]
		+	&= \int u\,du\\[5pt]
		+	&= \frac{1}{2}u^{2} + C\\[5pt]
		+	&= \frac{1}{2}\sin^2\!x + C\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

If we are to succeed in simplifying the integral with a substitution, we must find an expression \displaystyle u = u(x) so that the integral can be written as

\displaystyle \int \left(\begin{matrix} \text{something}\\ \text{in u} \end{matrix}\right)\cdot {u}'\,dx\,\textrm{.}

As our integral is written,

\displaystyle \int\sin x\cos x\,dx

we see that the second factor \displaystyle \cos x is a derivative of the first factor, \displaystyle \sin x. If \displaystyle u=\sin x, the integral can thus be written as

\displaystyle \int u\cdot u'\,dx

and this makes \displaystyle u=\sin x an appropriate substitution,

\displaystyle \begin{align} \int \sin x\cos x\,dx &= \left\{ \begin{align} u &= \sin x\\[5pt] du &= (\sin x)'\,dx = \cos x\,dx \end{align} \right\}\\[5pt] &= \int u\,du\\[5pt] &= \frac{1}{2}u^{2} + C\\[5pt] &= \frac{1}{2}\sin^2\!x + C\,\textrm{.} \end{align}