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 Post subject: Calculus (14)
PostPosted: Fri, 19 Dec 2008 23:31:49 UTC 
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Joined: Sat, 18 Mar 2006 08:42:24 UTC
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I thought, just for a change, we won't have a difficult problem this time. But there's something about this problem:

Find a short solution (as short as you can!) to this integral: $ \int \frac{dx}{\sin^3 x + \cos^3 x}. Have fun!


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PostPosted: Mon, 29 Dec 2008 23:14:02 UTC 
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cody wrote:
Hey Commutative:

It's been 10 days and no one has bit. May we please see your clever method?.

I started by noting it is a sum of cubes.


ok, here's a hint, or actually a solution:

Spoiler:
$ \frac{1}{ \sin^3 x + \cos^3 x }=\frac{2}{3}\left[\frac{1}{\sin x + \cos x} + \frac{\sin x + \cos x}{1 + (\sin x - \cos x)^2} \right]


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