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July 31, 2014

July 31, 2014

Posted by **Michael** on Thursday, April 12, 2012 at 4:52pm.

The integral from the square root of three over three to the square root of three of the function 6/(t^2+1)

- Calculus -
**Steve**, Thursday, April 12, 2012 at 5:22pmThis is one of your standard integrals:

∫dx/(x^2+1) = arctan(x)

now, what angle θ has tanθ = √3/3?

θ = pi/6

Now go and crank on it. If you get stuck, come on back.

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