Tính tích phân sau \(\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^3}\cos x\,dx} \) ta được:

Đặt \(\left\{ \begin{array}{l}u = {x^3}\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 3{x^2}dx\\v = \sin x\end{array} \right.\)

Khi đó ta có:

\(\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^3}\cos x\,dx}  \\= \left( {{x^3}\sin x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right. - 3\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {\sin x.{x^2}dx} \)

Đặt \(I = \int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^2}\sin x\,dx} \).            

Ta có: \(I = \int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^2}\sin x\,dx} \)\(\, = \left( { - {x^2}\cos x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right. + 2\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {\cos x.} \,xdx\)

Đặt \({I_1} = \int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {x\cos xdx} \)

Ta có: \({I_1} = \int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {x\cos xdx} \)\(\, = \left( {x\sin x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right. - \int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {\sin xdx} \)

\( = \left( {\dfrac{\pi }{3}.\dfrac{{\sqrt 3 }}{2} - \left( { - \dfrac{\pi }{3}} \right)\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right) - \left( { - \cos x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right.\)\( = 0 - \left( { - \dfrac{1}{2} - \left( { - \dfrac{1}{2}} \right)} \right) = 0\)

Khi đó \(I = \left( { - {x^2}\cos x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right. \)\(\,= \left( { - \dfrac{{{\pi ^2}}}{9}.\dfrac{1}{2}} \right) - \left( { - \dfrac{{{\pi ^2}}}{9}.\dfrac{1}{2}} \right) = 0\)

Khi đó \(\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^3}\cos x\,dx}\)\(\,  = \left( {{x^3}\sin x} \right)\left| {_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}}} \right. \)\(\,= \dfrac{{{\pi ^3}}}{{27}}.\dfrac{{\sqrt 3 }}{2} - \left( { - \dfrac{{{\pi ^3}}}{{27}}} \right)\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = 0\)

Chọn đáp án D.