Nếu \(\int\limits_0^1 {xf\left( x \right)d{\rm{x}}} = 4\) thì \(\int\limits_0^{\frac{\pi }{4}} {f\left( {{\rm{cos}}2{\rm{x}}} \right)}

Võ Chí Công

New member
Nếu \(\int\limits_0^1 {xf\left( x \right)d{\rm{x}}} = 4\) thì \(\int\limits_0^{\frac{\pi }{4}} {f\left( {{\rm{cos}}2{\rm{x}}} \right)} \sin 4{\rm{xdx}}\) bằng:
A. 2
B. 6
C. 8
D. 4
 

Nguyễn Huế

New member
Đặt \(t = \cos 2x \Rightarrow dt = - 2\sin 2xdx \Rightarrow \left\{ \begin{array}{l}x = 0,\,\,t = 1\\x = \frac{\pi }{4},\,\,t = 0\end{array} \right.\)
\( \Rightarrow \int\limits_0^{\frac{\pi }{4}} {f\left( {\cos 2{\rm{x}}} \right)\sin 4{\rm{xdx}} = - \int\limits_1^0 {tf\left( t \right)dt} } \)
\( \Rightarrow \int\limits_0^{\frac{\pi }{4}} {f\left( {\cos 2x} \right)\sin 4xdx = \int\limits_0^1 {tf\left( t \right)dt} = \int\limits_0^1 {xf\left( x \right)dx = 4.} } \)