[tex]\displaystyle I_1 = \int x^2\cos^2 x\,dx \\ I_2 = \int x^2\sin^2 x\,dx \\ \\ I_1+I_2 = \int x^2(\sin^2 x+\cos^2 x)\,dx = \int x^2\, dx = \dfrac{x^3}{3}+C \\ \\ I_1-I_2 = \int x^2(\cos^2 x-\sin^2 x)\,dx = \int x^2\cos2x\,dx = \\ \\ = \dfrac{1}{2}\int x^2(\sin2x)'\,dx = \dfrac{x^2\sin 2x}{2} - \int x\sin 2x\, dx = \\ \\ =\dfrac{x^2\sin 2x}{2}+\dfrac{1}{2}\int x(\cos 2x)'\, dx = \dfrac{x^2\sin 2x}{2}+\dfrac{x\cos 2x}{2}-\dfrac{1}{2}\int cos 2x \,dx=[/tex]
[tex]\displaystyle =\dfrac{x^2\sin 2x}{2}+\dfrac{x\cos 2x}{2}-\dfrac{\sin 2x}{4}+C \\ \\ \Rightarrow 2I_{1} = \dfrac{x^3}{3}+\dfrac{x^2\sin 2x}{2}+\dfrac{x\cos 2x}{2}-\dfrac{\sin 2x}{4}+C \\ \\ \Rightarrow I_{1} = \dfrac{x^3}{6}+\dfrac{x^2\sin 2x}{4}+\dfrac{x\cos 2x}{4}-\dfrac{\sin 2x}{8}+C[/tex]
[tex]\displaystyle V = \pi \int_0^{\frac{3\pi}{2}} I_{1}\, dx =\pi\Big(\dfrac{x^3}{6}+\dfrac{x^2\sin 2x}{4}+\dfrac{x\cos 2x}{4}-\dfrac{\sin 2x}{8}\Big)\Big|_{0}^{\frac{3\pi}{2}} =\\ \\ \pi \Big(\dfrac{27\pi^3}{48}+0-\dfrac{3\pi}{8}-0\Big) = \dfrac{9\pi^4}{16} - \dfrac{3\pi^2}{8}[/tex]
