a) \(\int\limits_1^2 {x{{(1 - x)}^5}dx} \)
Đặt \(t = 1 - x\)\( \Rightarrow dt = - dx\)
Đổi cận: \(x = 1 \Rightarrow t = 0\), \(x = 2 \Rightarrow t = - 1\)
Khi đó \(\int\limits_1^2 {x{{(1 - x)}^5}dx} \)\( = \int\limits_0^{ - 1} {\left( {1 - t} \right).{t^5}\left( { - dt} \right)} \) \( = \int\limits_{ - 1}^0 {\left( {{t^5} - {t^6}} \right)dt} \) \( = \left. {\left( {\dfrac{{{t^6}}}{6} - \dfrac{{{t^7}}}{7}} \right)} \right|_{ - 1}^0\) \( = 0 - \dfrac{1}{6} + \dfrac{{ - 1}}{7} = - \dfrac{{13}}{{42}}\)
b) \(\int\limits_0^{\ln 2} {\sqrt {{e^x} - 1} dx} \)
Đặt \(t = \sqrt {{e^x} - 1} \)\( \Rightarrow {t^2} = {e^x} - 1 \Rightarrow 2tdt = {e^x}dx\) \( \Rightarrow dx = \dfrac{{2tdt}}{{{e^x}}} = \dfrac{{2tdt}}{{{t^2} + 1}}\)
Đổi cận: \(x = 0 \Rightarrow t = 0\), \(x = \ln 2 \Rightarrow t = 1\).
Khi đó \(\int\limits_0^{\ln 2} {\sqrt {{e^x} - 1} dx} \)\( = \int\limits_0^1 {t.\dfrac{{2t}}{{{t^2} + 1}}dt} \) \( = \int\limits_0^1 {\left( {2 - \dfrac{2}{{{t^2} + 1}}} \right)dt} \) \( = 2 - 2\int\limits_0^1 {\dfrac{{dt}}{{{t^2} + 1}}} \)
Xét \(I = \int\limits_0^1 {\dfrac{{dt}}{{{t^2} + 1}}} \). Đặt \(t = \tan u \Rightarrow dt = \left( {1 + {{\tan }^2}u} \right)du\).
Đổi cận \(t = 0 \Rightarrow u = 0\), \(t = 1 \Rightarrow u = \dfrac{\pi }{4}\).
Khi đó \(I = \int\limits_0^1 {\dfrac{{dt}}{{{t^2} + 1}}} \)\( = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{1 + {{\tan }^2}u}}{{{{\tan }^2}u + 1}}du} = \int\limits_0^{\dfrac{\pi }{4}} {du} = \dfrac{\pi }{4}\)
Suy ra \(\int\limits_0^{\ln 2} {\sqrt {{e^x} - 1} dx} \)\( = 2 - 2.\dfrac{\pi }{4} = 2 - \dfrac{\pi }{2}\).
c) \(\int\limits_1^9 {x\sqrt[3]{{1 - x}}dx} \)
Đặt \(t = \sqrt[3]{{1 - x}}\) \( \Rightarrow {t^3} = 1 - x \Rightarrow 3{t^2}dt = - dx\)
Đổi cận \(x = 1 \Rightarrow t = 0\), \(x = 9 \Rightarrow t = - 2\)
Khi đó \(\int\limits_1^9 {x\sqrt[3]{{1 - x}}dx} \)\( = \int\limits_0^{ - 2} {\left( {1 - {t^3}} \right).t\left( { - 3{t^2}dt} \right)} \) \( = 3\int\limits_{ - 2}^0 {\left( {{t^3} - {t^6}} \right)dt} \) \( = 3\left. {\left( {\dfrac{{{t^4}}}{4} - \dfrac{{{t^7}}}{7}} \right)} \right|_{ - 2}^0\) \( = - 3\left( {\dfrac{{16}}{4} + \dfrac{{{2^7}}}{7}} \right) = - \dfrac{{468}}{7}\).
d) \(\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} \)
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 \(\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \dfrac{\pi }{2}.I = \dfrac{{{\pi ^2}}}{4}\).
e) \(\int\limits_{ - 1}^1 {{x^2}{{(1 - {x^3})}^4}dx} \)
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
\(\int\limits_{ - 1}^1 {{x^2}{{(1 - {x^3})}^4}dx} \)\( = \int\limits_2^0 {{t^4}.\left( { - \dfrac{{dt}}{3}} \right)} = \dfrac{1}{3}\int\limits_0^2 {{t^4}dt} \) \( = \dfrac{1}{3}\left. {\left( {\dfrac{{{t^5}}}{5}} \right)} \right|_0^2 = \dfrac{{32}}{{15}}\).
