Find the area of the regions by the lines and curves given below.

$\displaystyle y= 8cos(x)$ and $\displaystyle y = sec^2(x)$ between $\displaystyle \frac{-\pi}{3} \leq x \leq \frac{\pi}{3} $

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- May 17th 2008, 08:42 PMcasemeisterArea between two curves
Find the area of the regions by the lines and curves given below.

$\displaystyle y= 8cos(x)$ and $\displaystyle y = sec^2(x)$ between $\displaystyle \frac{-\pi}{3} \leq x \leq \frac{\pi}{3} $ - May 17th 2008, 08:53 PMo_O
First notice that the curves both intersect at $\displaystyle -\frac{\pi}{3}$ and $\displaystyle \frac{\pi}{3}$ which will be our limits of the integrand. Now, if you sketch it out, you'll see that $\displaystyle \cos x > \sec^{2} x$ over the interval $\displaystyle \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

So the area bounded is given by: $\displaystyle \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right) dx$

Should be simple to integrate. - May 17th 2008, 08:54 PMTheEmptySet
Here is a graph

Attachment 6362

$\displaystyle \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos(x)-\sec^{2}(x))dx=8\sin(x)-\tan(x)\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=6\sqrt{3}$ - May 17th 2008, 09:20 PMcasemeister
See, thats what I need some help on, integrating it. Im guessing you use Substitution, but i've tried that and i dont get anywhere, i just keep substituting multiple times.

- May 17th 2008, 09:22 PMo_O
$\displaystyle \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right)dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 8\cos x dx - \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec^{2} x dx$

You should know that: $\displaystyle \int \cos x dx = \sin x + C$

And: $\displaystyle \int sec^{2} x dx = \tan x + C$ - May 17th 2008, 09:24 PMMathstud28
- May 17th 2008, 09:29 PMcasemeister
Doh', I really should have listened to my professor and drill those derivatives into my head...