Cho hai hàm số \(f\left( x \right),g\left( x \right)\) liên tục trên \(\mathbb{R}\) và thỏa mãn \(f'\left( x \right) = \sin x,g'\left(

Cho hai hàm số \(f\left( x \right),g\left( x \right)\) liên tục trên \(\mathbb{R}\) và thỏa mãn \(f'\left( x \right) = \sin x,g'\left( x \right) = 2x\), \(f\left( {\frac{\pi }{2}} \right) = g\left( 0 \right) = 0\). Tính \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} \).
A. \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = - {x^2}\cos x + 2x\sin x + 2\cos x + C\)
B. \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = {x^2}\cos x - 2x\sin x + 2\cos x + C\)
C. \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = - {x^2}\cos x + 2x\sin x - 2\cos x + C\)
D. \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = {x^2}\cos x - 2x\sin x - 2\cos x + C\)

Ta có: \(\left\{ \begin{array}{l}f\left( x \right) = \int {\left( {\sin x} \right)dx} = - \cos x + C,f\left( {\frac{\pi }{2}} \right) = 0 \Rightarrow C = 0 \Rightarrow f\left( x \right) = - \cos x\\g\left( x \right) = \int {2xdx} = {x^2} + C,g\left( 0 \right) = 0 \Rightarrow C = 0 \Rightarrow g\left( x \right) = {x^2}\end{array} \right.\)
Suy ra \(\int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = \int {{x^2}\sin xdx} \)
Đặt \(\left\{ \begin{array}{l}u = {x^2}\\dv = \sin xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 2xdx\\v = - \cos x\end{array} \right. \Rightarrow \int {{x^2}\sin xdx} = - {x^2}\cos x + 2\int {x\cos xdx} \)
Đặt \(\left\{ \begin{array}{l}{u_1} = x\\d{v_1} = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}d{u_1} = dx\\{v_1} = \sin x\end{array} \right. \Rightarrow \int {{x^2}\sin xdx} = - {x^2}\cos x + 2x\sin x - 2\int {\sin xdx} \)
\( \Rightarrow \int {g\left( x \right)d\left[ {f\left( x \right)} \right]} = - {x^2}\cos x + 2x\sin x + 2\cos x + C.\)