Ta có: (I = intlimits_0^{frac{pi }{2}} {sin 2x.f’left( {sin x} right)dx} )
(= 2intlimits_0^{frac{pi }{2}} {sin x.cos x.f’left( {sin x} right)dx} ).
Đặt (t = sin x Rightarrow dt = cos xdx.)
Đổi cận: (left{ begin{array}{l}x = 0 Rightarrow t = 0\x = frac{pi }{2} Rightarrow t = 1end{array} right.).
Khi đó ta có: (I = 2intlimits_0^1 {t.f’left( t right)dt} ).
Đặt (left{ begin{array}{l}u = t\dv = f’left( t right)dtend{array} right. Rightarrow left{ begin{array}{l}du = dt\v = fleft( t right)end{array} right.).
(begin{array}{l} Rightarrow I = 2.left[ {left. {left( {t.fleft( t right)} right)} right|_0^1 – intlimits_0^1 {fleft( t right)dt} } right]\,,,,,,,,, = 2.left( {fleft( 1 right) – intlimits_0^1 {fleft( t right)dt} } right)\,,,,,,,,, = 2.left( {1 – frac{1}{2}} right) = 1.end{array})
Chọn D.