Question

Sol. [Key Concept: $\sin^{4}x + \cos^{4}x = 1 - \frac{1}{2}\sin^{2}(2x)$; integrate using the period over $20\pi$.]

$${I = \int_{0}^{20\pi}\mspace{2mu}\left( 1 - 2\sin^{2}x\cos^{2}x \right) \cdot dx }{= 20\pi - \frac{1}{2}\int_{0}^{20\pi}\mspace{2mu}\sin^{2}2xdx }{20\pi - \frac{1}{2}(40)\int_{0}^{\frac{\pi}{2}}\mspace{2mu}\sin^{2}2x }{20\pi - \frac{20}{2}\left( x - \frac{sin2x}{2} \right)_{0}^{\frac{\pi}{2}} }{20\pi - 10\left( \frac{\pi}{2} \right) = 15\pi}$$