Giá trị của tích phân

33/40

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

\[I = \frac{{32}}{{128}}\pi \]

\[I = \frac{{33}}{{128}}\pi \]

\[I = \frac{{31}}{{128}}\pi \]

\[I = \frac{{30}}{{128}}\pi \]

Giải thích

Ta có:

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

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

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

\[ = \left( {1 - \frac{1}{2}{{\sin }^2}2x} \right)\left( {1 - \frac{3}{4}{{\sin }^2}2x} \right) = 1 - \frac{5}{4}{\sin ^2}2x + \frac{3}{8}{\left( {{{\sin }^2}2x} \right)^2}\]

\[ = 1 - \frac{5}{4}.\frac{{1 - \cos 4x}}{2} + \frac{3}{8}.{\left( {\frac{{1 - \cos 4x}}{2}} \right)^2}\]

\[ = \frac{3}{8} + \frac{5}{8}\cos 4x + \frac{3}{{32}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right) = \frac{{15}}{{32}} + \frac{7}{{16}}\cos 4x + \frac{3}{{32}}{\cos ^2}4x\]

\[ = \frac{{15}}{{32}} + \frac{7}{{16}}\cos 4x + \frac{3}{{32}}.\frac{{1 + \cos 8x}}{2} = \frac{{33}}{{64}} + \frac{7}{{16}}\cos 4x + \frac{3}{{64}}\cos 8x\]

Do đó\[I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{33}}{{64}} + \frac{7}{{16}}\cos 4x + \frac{3}{{64}}\cos 8x} \right)dx\]

\[ = \frac{{33}}{{64}}x\left| {_0^{\frac{\pi }{2}}} \right. + \frac{7}{{64}}sin4x\left| {_0^{\frac{\pi }{2}}} \right. + \frac{3}{{512}}sin8x\left| {_0^{\frac{\pi }{2}}} \right. = = \frac{{33}}{{128}}\pi \]

Đáp án cần chọn là: B