What's Wrong With My Calculation of a Indefinite Integral?

Question

So, I was trying to find the integral of $$f(x)=\sin^{2}\left(\frac{33\pi}{x+\frac{1}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)}\right)\left(\frac{1}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)\right)+\left(\sin^{2}\left(\frac{33\pi}{x+\frac{3}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)}\right)\left(\frac{1}{2}-\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)\right)\right)$$ I tried to solve it by breaking it up into the two halves and integrate those, then add them back up: Then, I got: $$g(x)=\frac{\left(\arctan^{2}\left(\cot\left(\pi x\right)\right)-\pi^{2}x\right)\left(\cos\left(\frac{132\pi^{2}}{2\arctan\left(\cot\left(\pi x\right)\right)+2\pi x+\pi}\right)-1\right)}{4\pi^{2}}+\left(-\frac{\left(\arctan^{2}\left(\cot\left(\pi x\right)\right)+\pi^{2}x\right)\left(\cos\left(\frac{132\pi^{2}}{2\arctan\left(\cot\left(\pi x\right)\right)+\pi\left(2x+3\right)}\right)-1\right)}{4\pi^{2}}\right)$$ So, to test it, I put in $$\int_{0}^5 f(x) dx=2.90450849719$$ and $$g(5)-g(0)=3.97955872582$$ I'm asking, what did I do wrong?

Answer 1

Your function simplifies.

$$ f(x)=\sin^{2}\left(\frac{33\pi}{x+\frac{1}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)}\right)\left(\frac{1}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)\right)+\left(\sin^{2}\left(\frac{33\pi}{x+\frac{3}{2}+\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)}\right)\left(\frac{1}{2}-\frac{1}{\pi}\arctan\left(\cot\left(x\pi\right)\right)\right)\right)$$

Note that $$\arctan (\cot x\pi )=\pi/2 - x\pi $$

Therefore $$(1/\pi )(\arctan (\cot x\pi )=(1/\pi) (\pi/2 - x\pi)=1/2-x$$

Therefore your function simplifies $$ f(x) = \sin ^2(33\pi)(1-x)+\sin^2(33\pi/2)(1/2 +x) = 1/2 +x$$

Thus $$\int f(x) dx = x/2+x^2/2+C$$