shamieh
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Sketch the region bounded by the line y = 2 and the graph of $$y = sec^2(3x)$$ for $$ \frac{-\pi}{6} < x < \frac{\pi}{6}$$
Very very confused on this problem...
So here is what I set up..
Top function - bottom function
$$\int ^{\pi/6}_{-\pi/6} 2 - \sec^2(3x) dx$$
so I split the integrals in two.
for the first integral, i let u = 3x , du/3 = dx
thus:
$$1/3 \int ^{\pi/6}_{-\pi/6} sec^2(u) du$$
which = $$1/3tan(u)$$ ... Updating the limits $$|^{\pi/2} _ 0$$
here is where I am running into a problem... $$tan(\pi/2)$$ doesn't exist it is infinity or undefined but I'm bounded by y = 2...So why am i running into this problem? If i am bounded then won't i still end up with 2 - infinity? I am confused. Maybe I'm approaching this the wrong way.
Very very confused on this problem...
So here is what I set up..
Top function - bottom function
$$\int ^{\pi/6}_{-\pi/6} 2 - \sec^2(3x) dx$$
so I split the integrals in two.
for the first integral, i let u = 3x , du/3 = dx
thus:
$$1/3 \int ^{\pi/6}_{-\pi/6} sec^2(u) du$$
which = $$1/3tan(u)$$ ... Updating the limits $$|^{\pi/2} _ 0$$
here is where I am running into a problem... $$tan(\pi/2)$$ doesn't exist it is infinity or undefined but I'm bounded by y = 2...So why am i running into this problem? If i am bounded then won't i still end up with 2 - infinity? I am confused. Maybe I'm approaching this the wrong way.