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} $
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} $
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.
$\displaystyle \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 8\cos x - \sec^{2} x \right)dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 8\cos x dx - \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \sec^{2} x dx$
You should know that: $\displaystyle \int \cos x dx = \sin x + C$
And: $\displaystyle \int sec^{2} x dx = \tan x + C$