(i)[3]
Show that $12\sin^2 x\cos^2 x = \frac{3}{2}(1 - \cos 4x)$.
(ii)[3]
Hence show that $\displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 12\sin^2 x\cos^2 x\,dx = \frac{\pi}{8} + \frac{3\sqrt{3}}{16}$.
Mathematics 9709 · AS & A Level · Integration
Show that $12\sin^2 x\cos^2 x = \frac{3}{2}(1 - \cos 4x)$.
Hence show that $\displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 12\sin^2 x\cos^2 x\,dx = \frac{\pi}{8} + \frac{3\sqrt{3}}{16}$.