Calculus 1 - Find the area of shaded region
Sophia Terry 
I have been working at this for like 3 hours and can find nothing online like it and there are no examples in the text book. I must be missing a simple step somewhere.
The correct answer is given as $$\frac{3 \pi \sqrt 2}{2}$$ but I cannot figure out how it was obtained.
Thank you for any help.
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$\begingroup$Hint $\#1$: (calculus way)
The area you are trying to find is between the graphs
$$f(x) = 3 \sqrt 2 \qquad g(x) = 3 \sec x \tan x$$
on the interval $(-\pi/4, \pi/4)$.
(I don't know why they expressed the latter function in terms of $\theta$, but whatever.)
The area is then given by
$$\text{shaded area} = \int_{-\pi/4}^{\pi/4} \Big( f(x) - g(x) \Big) \, dx$$
(In case the thing regarding $\theta$ in lieu of $x$ seems potentially an issue, WolframAlpha does happen to give the correct answer, if you just note that $\sqrt 2/2 = 1/\sqrt 2$.)
Hint $\#2$: (geometric way)
You should be able to convince yourself that the blue area makes up half of the area of the rectangle the picture defines:
(The side lengths of this rectangle are $\pi/2$ and $6 \sqrt 2$.) So if you can somehow convince yourself that the blue area is half, you can avoid calculus altogether!
(This can be somewhat justified by noticing that $y(x) = 3 \sec x \tan x$ is an odd function, i.e. $y(-x) = -y(x)$, and we are over an interval symmetric about zero.)
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