Giá trị của tích phân $I = \int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^6}x + {{\cos }^6}x} \right)dx}$ là:

Ta có:

$\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$

$= \left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right]$$\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]$

$= \left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)\left( {1 - \dfrac{3}{4}{{\sin }^2}2x} \right)$$= 1 - \dfrac{5}{4}{\sin ^2}2x + \dfrac{3}{8}{\left( {{{\sin }^2}2x} \right)^2}$$= 1 - \dfrac{5}{4}.\dfrac{{1 - \cos 4x}}{2} + \dfrac{3}{8}.{\left( {\dfrac{{1 - \cos 4x}}{2}} \right)^2}$$= \dfrac{3}{8} + \dfrac{5}{8}\cos 4x + \dfrac{3}{{32}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)$$= \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}{\cos ^2}4x$$= \dfrac{{15}}{{32}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{32}}.\dfrac{{1 + \cos 8x}}{2}$$= \dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x$

Do đó $I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\dfrac{{33}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{3}{{64}}\cos 8x} \right)dx}$$= \left. {\dfrac{{33}}{{64}}x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{7}{{64}}\sin 4x} \right|_0^{\dfrac{\pi }{2}} + \left. {\dfrac{3}{{512}}\sin 8x} \right|_0^{\dfrac{\pi }{2}}$$= \dfrac{{33}}{{128}}\pi$