Answer:
[tex]=\pi[/tex]
Step-by-step explanation:
[tex]\int _0^{2\pi }\sin ^2\left(x\right)-\cos ^3\left(x\right)dx[/tex]
compute the indefinite integral
[tex]\int \sin ^2\left(x\right)-\cos ^3\left(x\right)dx=\frac{1}{2}x-\frac{1}{4}\sin \left(2x\right)-\sin \left(x\right)+\frac{1}{3}\sin ^3\left(x\right)+C\\\int \sin ^2\left(x\right)-\cos ^3\left(x\right)dx\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\int \sin ^2\left(x\right)dx=\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)\\\int \cos ^3\left(x\right)dx=\sin \left(x\right)-\frac{\sin ^3\left(x\right)}{3}[/tex]
[tex]=\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)-\left(\sin \left(x\right)-\frac{\sin ^3\left(x\right)}{3}\right)\\\mathrm{Simplify\:}\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)-\left(\sin \left(x\right)-\frac{\sin ^3\left(x\right)}{3}\right):\quad \frac{1}{2}x-\frac{1}{4}\sin \left(2x\right)-\sin \left(x\right)+\frac{1}{3}\sin ^3\left(x\right)\\=\frac{1}{2}x-\frac{1}{4}\sin \left(2x\right)-\sin \left(x\right)+\frac{1}{3}\sin ^3\left(x\right)\\Add\:a\:constant\:to\:the\:solution[/tex]
[tex]=\frac{1}{2}x-\frac{1}{4}\sin \left(2x\right)-\sin \left(x\right)+\frac{1}{3}\sin ^3\left(x\right)+C\\\mathrm{Compute\:the\:boundaries}:\quad \int _0^{2\pi }\sin ^2\left(x\right)-\cos ^3\left(x\right)dx=\pi -0\\=\pi -0\\\mathrm{Simplify}\\=\pi[/tex]
