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October 2, 2014

October 2, 2014

Posted by **Agnes** on Thursday, March 20, 2008 at 12:43pm.

can someone help me start his problem?

- Calculus -
**drwls**, Thursday, March 20, 2008 at 12:51pmInterval, or integral?

I suggest you split it up into

4 sec^2 (theta)

+ 4 sin(theta)/cos^2(theta)

- Calculus -
**drwls**, Thursday, March 20, 2008 at 6:38pmYou can find the integral of the first one in a table of integrals. The second one is easy if you let u = cos theta and du = -sin theta dtheta

- Calculus -
**drwls**, Thursday, March 20, 2008 at 8:11pmThe integral of sec^2 is tan

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