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Lösung 2.1:3b

Aus Online Mathematik Brückenkurs 2

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As the integral stands, it is not so easy to see what the primitive functions are, but if we use the formula for double angles,
As the integral stands, it is not so easy to see what the primitive functions are, but if we use the formula for double angles,
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{{Displayed math||<math>\int 2\sin x\cos x\,dx = \int \sin 2x\,dx</math>}}
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<math>\int{2\sin x\cos x}\,dx=\int{\sin 2x}\,dx</math>
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we obtain a standard integral where we can write down the primitive functions directly,
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{{Displayed math||<math>\int \sin 2x\,dx = -\frac{\cos 2x}{2}+C\,,</math>}}
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we obtain a standard integral where we can write down the primitive functions directly:
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where <math>C</math> is an arbitrary constant.
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<math>\int{\sin 2x}\,dx=-\frac{\cos 2x}{2}+C</math>
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where
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<math>C</math>
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is an arbitrary constant.
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Version vom 13:13, 21. Okt. 2008

As the integral stands, it is not so easy to see what the primitive functions are, but if we use the formula for double angles,

2sinxcosxdx=sin2xdx 

we obtain a standard integral where we can write down the primitive functions directly,

sin2xdx=−2cos2x+C 

where C is an arbitrary constant.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:3b“