Strange definite integral

Question

So I have this integral $$\int_0^{2\pi}\mathrm{arccot}(\cot(f))\cos(t)\cdot21\,\mathrm df$$ and it's solution in indefinite form is $$\begin{align} \int21\cos(t)\mathrm{arccot}(\cot(f))\,\mathrm df &= 21\cos(t)f\mathrm{arccot}(\cot(f)) - \dfrac{21\cos(t)f^2}2 + C \\ &=\dfrac{21\cos(t)f^2}2 + C \end{align}$$ Strange thing is that if I try to plug in (for the definite integral), $2\pi$ in the not simplified form, i.e., $$21\cos(t)f\mathrm{arccot}(\cot(f)) - \dfrac{21\cos(t)f^2}2 + C,$$ I get $\cot(2\pi)$ which is undefined. But if I plug the limits in the simplified form of the indefinite integral, i.e., $$\dfrac{21\cos(t)f^2}2 + C,$$I get the result $$42\pi^2\cos(t)$$ So my question is whether the calculator is right by plugging in the limits in the simplified form, and also if it is right, how can I get two different answers for the same definite integral?

Answer 1

I would write $$21\cdot \cot(t)\int arccot(\cot(f))df$$ and use integration by parts.