Đặt \(t = \sqrt x \Rightarrow {t^2} = x\) \( \Rightarrow 2tdt = dx\)
Khi đó \(\int\limits_0^1 {\sin \sqrt x dx} \)\( = \int\limits_0^1 {\sin t.2tdt} = 2\int\limits_0^1 {t\sin tdt} \)
Đặt \(\left\{ \begin{array}{l}u = t\\dv = \sin tdt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = - \cos t\end{array} \right.\)
\( \Rightarrow \int\limits_0^1 {t\sin tdt} = \left. { - t\cos t} \right|_0^1 + \int\limits_0^1 {\cos tdt} \) \( = - 1\cos 1 + \left. {\sin t} \right|_0^1 = - \cos 1 + \sin 1\)
Vậy \(\int\limits_0^1 {\sin \sqrt x dx} = 2\left( {\sin 1 - \cos 1} \right)\)
Chọn A.
