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Lösung 2.1:5b

Aus Online Mathematik Brückenkurs 2

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In this case, we can use the formula for half-angles and write
In this case, we can use the formula for half-angles and write
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{{Displayed math||<math>\sin^2\!x=\frac{1-\cos 2x}{2}\,\textrm{.}</math>}}
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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
The right-hand side contains only terms which we can integrate and the calculation becomes
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{{Displayed math||<math>\begin{align}
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\int \sin^2\!x\,dx
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&= \int\frac{1-\cos 2x}{2}\,dx\\[5pt]
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&= \int\Bigl(\frac{1}{2}-\frac{1}{2}\cos 2x\Bigr)\,dx\\[5pt]
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&= \frac{x}{2} - \frac{1}{2}\cdot\frac{\sin 2x}{2} + C\\[5pt]
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&= \frac{x}{2} - \frac{\sin 2x}{4} + C\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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(Notice how we compensate with a factor 2 in the denominator for the inner derivative of <math>\sin 2x\,</math>.)
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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>.)
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Version vom 10:54, 28. 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

sin2xdx=21−cos2xdx=21−21cos2xdx=x2−212sin2x+C=x2−4sin2x+C.

(Notice how we compensate with a factor 2 in the denominator for the inner derivative of sin2x.)

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