Toán 12 Tính tích phân

Cho tích phân \(I = \int\limits_0^\pi {{{\sin }^2}x.{{\cos }^2}xdx.}\) Khẳng định nào sau đây là khẳng định đúng?
A. \(I = \frac{1}{4}\left( {\frac{1}{2}x - \frac{1}{4}\sin 4x} \right)\left| {\begin{array}{*{20}{c}} {^\pi }\\ {_0} \end{array}} \right.\)
B. \(I = \frac{1}{8}\left( {x + \frac{1}{4}\sin 4x} \right)\left| {\begin{array}{*{20}{c}} {^\pi }\\ {_0} \end{array}} \right.\)
C. \(I = \frac{1}{8}\left( {x - \frac{1}{4}\sin 4x} \right)\left| {\begin{array}{*{20}{c}} {^\pi }\\ {_0} \end{array}} \right.\)
D. \(I = \frac{1}{4}\left( {\frac{1}{2}x + \frac{1}{4}\sin 4x} \right)\left| {\begin{array}{*{20}{c}} {^\pi }\\ {_0} \end{array}} \right.\)

\(I = \int\limits_0^\pi {{{\sin }^2}x.{{\cos }^2}xdx} = \frac{1}{4}\int\limits_0^\pi {{{\sin }^2}2xdx}\)
\(= \frac{1}{8}\int\limits_0^\pi {\left( {1 - \cos 4x} \right)dx} = \frac{1}{8}\left( {x - \sin 4x} \right)\left| {\begin{array}{*{20}{c}} {^\pi }\\ {_0} \end{array}} \right.\)