Правильно ли решён интеграл?

4 - x^2 = 4*(1 - x^2/4) = 4*(1 - (x/2)^2 ) x/2 = sin t; x = 2sin t; dx = 2cos t dt; 4*(1 - (x/2)^2 ) = 4cos^2 t Пределы интегрирования: x1 = -2 = 2sin t; sin t = -1; t = -pi/2 x2 = 2 = 2sin t; sin t = 1; t = pi/2 $\int\limits^{\pi/2}_{-\pi/2} {\sqrt{(4cos^2(t))^3}*2cos(t)} \, dt =\int\limits^{\pi/2}_{-\pi/2} {(2cos(t))^3*2cos(t)} \, dt=\\ =16\int\limits^{\pi/2}_{-\pi/2} {cos^4(t)} \, dt=16\int\limits^{\pi/2}_{-\pi/2} {(\frac{1}{2}(cos(2t)+1))^2} \, dt =$ $=4\int\limits^{\pi/2}_{-\pi/2} {(cos^2(2t)+2cos(2t)+1)} \, dt=\\=4\int\limits^{\pi/2}_{-\pi/2} {(\frac{1}{2}*(cos(4t)+1)+2cos(2t)+1)} \, dt=\\ =\int\limits^{\pi/2}_{-\pi/2} {(2cos(4t)+8cos(2t)+6)} \, dt=(\frac{2}{4}sin(4t) +\frac{8}{2}sin(2t)+6t)|^{\pi/2}_{-\pi/2}$ $=\frac{1}{2}(sin(2\pi)-sin(-2\pi))+4(sin(\pi)-sin(-\pi))+6(\frac{\pi}{2}-(-\frac{\pi}{2}))=\\ =\frac{1}{2}(0-0)+4(0-0)+6\pi = 6\pi$