Bài tập 3.17 trang 170 SBT Toán 12

\(\int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} \)

Đặt \(x = \pi  - t\), ta suy ra:

\(\int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} \)\( = \int\limits_\pi ^0 {\dfrac{{\left( {\pi  - t} \right)\sin \left( {\pi  - t} \right)}}{{1 + {{\cos }^2}\left( {\pi  - t} \right)}}\left( { - dt} \right)} \)\( = \int\limits_0^\pi  {\dfrac{{\left( {\pi  - t} \right)\sin t}}{{1 + {{\cos }^2}t}}dt} \) \( = \int\limits_0^\pi  {\dfrac{{\left( {\pi  - x} \right)\sin x}}{{1 + {{\cos }^2}x}}dx} \)

\( = \int\limits_0^\pi  {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}dx}  - \int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} \)

\( \Rightarrow 2\int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}  = \int\limits_0^\pi  {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}dx} \) \( \Leftrightarrow \int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}  = \dfrac{\pi }{2}\int\limits_0^\pi  {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} \)

Xét \(I = \int\limits_0^\pi  {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} \), đặt \(u = \cos x \Rightarrow du =  - \sin xdx\).

Đổi cận \(x = 0 \Rightarrow u = 1,\) \(x = \pi  \Rightarrow u =  - 1\). Khi đó

\(I = \int\limits_0^\pi  {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}dx} \)\( = \int\limits_1^{ - 1} {\dfrac{{ - du}}{{1 + {u^2}}}}  = \int\limits_{ - 1}^1 {\dfrac{{du}}{{1 + {u^2}}}} \)

Đặt \(u = \tan t\) \( \Rightarrow du = \left( {1 + {{\tan }^2}t} \right)dt\). Đổi cận \(u =  - 1 \Rightarrow t =  - \dfrac{\pi }{4},\) \(u = 1 \Rightarrow t = \dfrac{\pi }{4}\).

Khi đó \(I = \int\limits_{ - 1}^1 {\dfrac{{du}}{{1 + {u^2}}}}  = \int\limits_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {dt}  = \dfrac{\pi }{2}\)

Vậy \(\int\limits_0^\pi  {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}  = \dfrac{\pi }{2}.I = \dfrac{{{\pi ^2}}}{4}\).

\(\int\limits_{ - 1}^1 {{x^2}{{(1 - {x^3})}^4}dx} \)

Đặt \(t = 1 - {x^3} \Rightarrow dt =  - 3{x^2}dx\) \( \Rightarrow {x^2}dx =  - \dfrac{{dt}}{3}\).

Đổi cận \(x =  - 1 \Rightarrow t = 2\), \(x = 1 \Rightarrow t = 0\). Khi đó

\(\int\limits_{ - 1}^1 {{x^2}{{(1 - {x^3})}^4}dx} \)\( = \int\limits_2^0 {{t^4}.\left( { - \dfrac{{dt}}{3}} \right)}  = \dfrac{1}{3}\int\limits_0^2 {{t^4}dt} \) \( = \dfrac{1}{3}\left. {\left( {\dfrac{{{t^5}}}{5}} \right)} \right|_0^2 = \dfrac{{32}}{{15}}\).