Tính tích phân \(\int\limits_{ - \dfrac{\pi }{3}}^{\dfrac{\pi }{3}} {{x^3}\cos x\,dx} \) ta được:
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Lời giải:
Báo saiĐặt \(\left\{ \begin{array}{l}u = {x^3}\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 3{x^2}dx\\v = \sin x\end{array} \right.\)
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.
