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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 12:50:39 IST
Accepted Answer [?]
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sinx.sin2x.sin3x=sin4x/4-sin6x/4+sin2x/4[use trignometric transformations].Hence it's integral is
-cos4x/16+cos6x/24-cos2x/8=(1/48)(2cos6x-3cos4x-6cos2x)
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MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
this reply: 7 points
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