From Förberedande kurs i matematik 2

Revision as of 14:40, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.1:5b moved to Solution 2.1:5b: Robot: moved page) ← Previous diff		Current revision (10:54, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The integral is not in our list of known integrals, but we will try to rewrite the integrand as something more manageable.
-	<center> [[Image:2_1_5b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	In this case, we can use the formula for half-angles and write
		+
		+	{{Displayed math||<math>\sin^2\!x=\frac{1-\cos 2x}{2}\,\textrm{.}</math>}}
		+
		+	The right-hand side contains only terms which we can integrate and the calculation becomes
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int \sin^2\!x\,dx
		+	&= \int\frac{1-\cos 2x}{2}\,dx\\[5pt]
		+	&= \int\Bigl(\frac{1}{2}-\frac{1}{2}\cos 2x\Bigr)\,dx\\[5pt]
		+	&= \frac{x}{2} - \frac{1}{2}\cdot\frac{\sin 2x}{2} + C\\[5pt]
		+	&= \frac{x}{2} - \frac{\sin 2x}{4} + C\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	(Notice how we compensate with a factor 2 in the denominator for the inner derivative of <math>\sin 2x\,</math>.)

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Current revision

The integral is not in our list of known integrals, but we will try to rewrite the integrand as something more manageable.

In this case, we can use the formula for half-angles and write

\displaystyle \sin^2\!x=\frac{1-\cos 2x}{2}\,\textrm{.}

The right-hand side contains only terms which we can integrate and the calculation becomes

\displaystyle \begin{align} \int \sin^2\!x\,dx &= \int\frac{1-\cos 2x}{2}\,dx\\[5pt] &= \int\Bigl(\frac{1}{2}-\frac{1}{2}\cos 2x\Bigr)\,dx\\[5pt] &= \frac{x}{2} - \frac{1}{2}\cdot\frac{\sin 2x}{2} + C\\[5pt] &= \frac{x}{2} - \frac{\sin 2x}{4} + C\,\textrm{.} \end{align}

(Notice how we compensate with a factor 2 in the denominator for the inner derivative of \displaystyle \sin 2x\,.)