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this is my question prompt.

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

$y=4\cos(x)$
$y=(6\sec(x))^2$
$x=\pi/4$
$x=-\pi/4$

Help and step by step explanations would be helpful! I graphed it however they don't intersect how am I going to find the area between these two equations?

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"I graphed it however they don't intersect"

Half true. Here, you're given four curves: $y = 4\cos{x}$, $y=(6\sec{x})^2$, $x = -\pi/4$, $x = \pi/4$.

What you probably did was graph the first two and conclude that they don't intersect anywhere.

But remember, you have four curves. The two curves that are stated as $x = $ something are vertical lines in your $xy$-plane. So, if you graph a vertical line at these two values of $x$, you'll find four total intersection points.

These four points bound a region: on the left and right we're bounded by our values for $x$, and on the top and bottom we're bounded by our equations for $y$. (The other answer posted has a nice picture showing the four intersection points - I was in the process of making a figure myself, but referring to that one is easier.)

So, when you set up your integral, you should be integrating over $x$ with the bounds specified, and your integrand should be the difference in functional value:

$$\int_{-\pi/4}^{\pi/4}{(6\sec{x})^2 - 4\cos{x}}\ dx$$

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Your statement about no intersections is not true, look closer:

region

There is an area bounded by the graphs of the four equations.

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