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Lösung 2.1:5b

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.1:5b moved to Solution 2.1:5b: Robot: moved page)
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The integral is not in our list of known integrals, but we will try to rewrite the integrand as something more manageable.
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<center> [[Image:2_1_5b.gif]] </center>
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In this case, we can use the formula for half-angles and write
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<math>\sin ^{2}x=\frac{1-\cos 2x}{2}</math>
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The right-hand side contains only terms which we can integrate and the calculation becomes
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<math>\begin{align}
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& \int{\sin ^{2}x\,dx}=\int{\frac{1-\cos 2x}{2}\,dx} \\
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& =\int{\left( \frac{1}{2}-\frac{1}{2}\cos 2x \right)}\,dx \\
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& =\frac{x}{2}-\frac{1}{2}\centerdot \frac{\sin 2x}{2}+C \\
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& =\frac{x}{2}-\frac{\sin 2x}{4}+C \\
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\end{align}</math>.
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(Notice how we compensate with a factor
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<math>\text{2}</math>
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in the denominator for the derivative of
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<math>\sin 2x</math>.)

Version vom 08:26, 19. Okt. 2008

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


sin2x=21−cos2x

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} \\ & =\int{\left( \frac{1}{2}-\frac{1}{2}\cos 2x \right)}\,dx \\ & =\frac{x}{2}-\frac{1}{2}\centerdot \frac{\sin 2x}{2}+C \\ & =\frac{x}{2}-\frac{\sin 2x}{4}+C \\ \end{align}.

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

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5b“