# Math Help - Area between two curves

1. ## Area between two curves

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

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

2. First notice that the curves both intersect at $-\frac{\pi}{3}$ and $\frac{\pi}{3}$ which will be our limits of the integrand. Now, if you sketch it out, you'll see that $\cos x > \sec^{2} x$ over the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

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

Should be simple to integrate.

3. Originally Posted by casemeister
Find the area of the regions by the lines and curves given below.

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

$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos(x)-\sec^{2}(x))dx=8\sin(x)-\tan(x)\bigg|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=6\sqrt{3}$

4. 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.

5. $\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right)dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 8\cos x dx - \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec^{2} x dx$

You should know that: $\int \cos x dx = \sin x + C$
And: $\int sec^{2} x dx = \tan x + C$

6. Originally Posted by casemeister
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.
$\int{8\cos(x)-sec^{2}(x)dx}$

seperating we get $\int{8\cos(x)dx}-\int{\sec^2(x)dx}$

Note that $\frac{D[\tan(x)]}{dx}=\sec^2(x)$

So $\int{8\cos(x)dx}-\int{\sec^2(x)dx}=8\sin(x)-\tan(x)+C$

7. Doh', I really should have listened to my professor and drill those derivatives into my head...