One must assume that shipgirls would need to do all the plotting of firing solutions though - meaning reasonably complex maths. (Unless they have mathfaries to do all that for them!)
Funny thing is that I actually learned this back in high school and I remember it as not being that hard. Two thing for sure are that I don't remember how I got over them all (I was bad at other parts of high school math) and I can't, for the life of me do math anymore now, ie using calculator for even easy maths.
Calculus is still easier than Probability and Statistics (AKA "Sadistics"), at least... (I remember half my class failed those, and I just BARELY survived through the use of every single TA availability given to us. I seriously spent three times as much time on those classes than any other class... The teacher was reprimanded for how many of her students were failing, though, so it may have partially been her...) That's even Single-Variable Calculus, which is the first year/easiest of it. They tend to look a lot scarier than they actually are. (You just add or subtract a power and divide or multiply by the value of the power, depending on whether it's integrals or derivatives.) As Demundo mentions, that's stuff they'll even teach in High School.
And yes, calculus is definitely constantly required for anything involving physics, including calculation of cannonfire trajectories, as seen here.
My high school gave us the option of Calculus and/or Statistics in the last two years (or applied mathematics if you were mathematically impaired); and for (one of) my degrees, we had to choose two of Maths (read calculus), Statistics and/or Information Systems. Didn't need Maths for my major, so I noped out of that one.
So really, I managed to completely avoid anything but the most basic calculus!
My biggest regret of high school and having taken Calculus but not having passed it due to senioritis. This was during the time I spent playing games 12 hours a day after coming home from school.
Actually, those kind of equation could be done by a calculator. Seriously, FX could solve this.
with trigonometry? Calculator could do that!? back in my high school day, trigonometry and logarithm is the bane of my existence (now it's the maths itself).
To be honest though I was actually pretty good at math supposedly, considering I passed ap calc in high school. But I can't do anything above simple math for the life of me now lol.
Find the definite integral of sin(x)/(sin^2(x)+1), with respect to x, from 0 to pi, and multiply by pi/2. Deceptively simple, and then you run into partial fraction decomposition huehuehue.
(Notation's gonna be a bit weird since I don't think Danbooru has LaTeX formatting.)
(I'm also going to omit the +C for when we solve indefinite integrals, because this is overall a definite integral and the +C, y'know, goes away when you solve a definite integral.)
2. u-substitution, which everyone hates (not as much as I have uv substitution, though): Let u = cos(x), du/dx = -sin(x) -> du = -sin(x)dx, therefore integral(-sin(x)dx/(cos^2(x)-2)) = integral(du/(u^2-2))
3. Time for algebra; the denominator is actually a factor-able polynomial: integral(du/(u^2-2)) = integral(du/((u-sqrt(2))(u+sqrt(2)))
5. Fun fact: integrals are actually linear operations; abuse this fact: integral(du((1/(2^(3/2)(u-sqrt(2)))) - (1/(2^(3/2)(u+sqrt(2)))))) = (1/2^(3/2))*(integral(du/(u-sqrt(2))) - (1/2^(3/2))*integral(du/(u+sqrt(2))) (don't forget we have a "multiply the whole damn thing by pi/2" floating outside of this integration, I've just been too lazy to keep track of it)
6. Now we have two integrals to solve! Thankfully, they are easier. Let's solve integral(du/(u-sqrt(2))) first:
6.1. More substitution - let's use v to disambiguate it from u: let v = u-sqrt(2), dv/du = 1 (remember, derivation is also a linear operation, and the derivative of any constant value is 0) -> dv = du, therefore integral(du/(u-sqrt(2))) = integral(dv/v)
6.2. integral(dv/v) is a known integral that evaluates to something sane: integral(dv/v) = ln(v)
6.3. Undo the substitution of v = u-sqrt(2): ln(v) = ln(u-sqrt(2))
7. Now we solve integral(du/(u+sqrt(2))):
7.1-7.3 Fun fact, this is literally exactly the same as 6.1-6.3, except that there's a + sign instead of a - sign; therefore integral(du/u+sqrt(2)) = ln(u+sqrt(2))
8. Plug it in, plug it in: (1/2^(3/2))*(integral(du/(u-sqrt(2))) - (1/2^(3/2))*integral(du/(u+sqrt(2))) = ln(u-sqrt(2))/2^(3/2) - ln(u+sqrt(2))/2^(3/2)
9. Undo the substitution of u = cos(x): ln(u-sqrt(2))/2^(3/2) - ln(u+sqrt(2))/2^(3/2) = ln(cos(x)-sqrt(2))/2^(3/2)) - ln(cos(x)+sqrt(2))/2^(3/2))
12. Solving for the definite integral from 0 to pi: (ln(sqrt(2)-cos(pi)) - ln(cos(pi)+sqrt(2)))/2^(3/2) - (ln(sqrt(2)-cos(0)) - ln(cos(0)+sqrt(2)))/2^(3/2)
12.1. Remember that cos(pi) = -1 and cos(0) = 1: (ln(sqrt(2)-(-1)))-ln(-1+sqrt(2)))/2^(3/2) - (ln(sqrt(2)-1)-ln(1+sqrt(2)))/2^(3/2) = (ln(sqrt(2)+1)-(ln(sqrt(2)-1)))/2^(3/2) - (ln(sqrt(2)-1)-ln(sqrt(2)+1))/2^(3/2)
12.2. Don't forget to multiply this whole thing by pi/2 as stated in the original formulation of the problem: (pi/2)((ln(sqrt(2)+1)-(ln(sqrt(2)-1)))/2^(3/2) - (ln(sqrt(2)-1)-ln(sqrt(2)+1))/2^(3/2))
12.3. At this point, you're done. The exact result is that line above. You can simplify it down to pi*tanh^-1(1/sqrt(2))/sqrt(2), though (which is I think transcendental, so you're not getting a nice decimal answer), or approximate it to ~1.9579 with a calculator.
tl;dr the answer is "almost 2".
(I won't lie, I resorted to a calculator at step 10, since I totally blanked on the whole "ln(x) shits the bed when x<0, what do" bit. Also the partial fraction decomp took me a bit, I always have trouble recognizing when to do it...)
