Stuck on trying to integrate

(4 + 3sec(2x))^2

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- Dec 24th 2008, 05:48 AMBeardIntegration
Stuck on trying to integrate

(4 + 3sec(2x))^2 - Dec 24th 2008, 06:23 AMgalactus
If we expand it out, we get:

$\displaystyle \int\left[16+24sec(2x)+9sec^{2}(2x)\right]dx$

Now, integrate term by term and it ain't bad. - Dec 24th 2008, 01:28 PMBeard
- Dec 24th 2008, 01:42 PMJameson