\int _0^{2\pi }\:\sqrt{1-t^2}dt

asked by guest
on Mar 23, 2025 at 12:09 pm



You asked:

Evaluate the integral: ∫02π1−t2dt\int_{0}^{2 \pi}{\sqrt{1 - {t}^{2}}}dt

MathBot Answer:

Definite integral

∫02π1−t2dt=arcsin⁡(2π)2+π1−4π2\int_{0}^{2 \pi}{\sqrt{1 - {t}^{2}}}dt = \frac{\arcsin{\left(2 \pi \right)}}{2} + \pi \sqrt{1 - 4 \pi^{2}}

+ Submit Question