Giải bài 3.17 trang 170 - SBT Đại số và Giải tích lớp 12
Tính các tích phân sau bằng phương pháp đổi biến số:
a) \(\int\limits_{1}^{2}{x\left( 1-x \right)^5dx}\) (đặt \(t=1-x\))
b) \(\int\limits_{0}^{ln 2}{\sqrt{e^x-1}dx}\) (đặt \(t=\sqrt{e^x-1}\))
c) \(\int\limits_{1}^{9}{x\sqrt[3]{1-x}dx}\) (đặt \(t=\sqrt[3]{1-x}\))
d) \(\int\limits_{0}^{\pi}{\dfrac{x \sin x}{1+\cos^2 x}dx}\) (đặt \(x=\pi -t\))
e) \(\int\limits_{-1}^{1}{x^2\left( 1-x^3 \right)^4dx}\)
(Đề thi tốt nghiệp THPT năm 2008)
a) Đặt \(t=1-x \Rightarrow dx=-dt\)
\(x=1\Rightarrow t=0\\x=2\Rightarrow t=-1\)
Ta có:
\(\begin{aligned} & \int\limits_{0}^{-1}{\left( 1-t \right){{t}^{5}}\left( -dt \right)}=\int\limits_{-1}^{0}{\left( {{t}^{5}}-{{t}^{6}} \right)dt} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{1}{6}{{t}^{6}}-\frac{1}{7}{{t}^{7}} \right)\left| \begin{aligned} & 0 \\ & -1 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{1}{6}-\frac{1}{7} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{13}{42} \\ \end{aligned} \)
b) Đặt \(t=\sqrt{e^x-1}\Rightarrow t^2=e^x-1\Rightarrow 2tdt=e^xdx=(t^2+1)dx \Rightarrow dx=\dfrac{2tdt}{t^2+1}\)
\(x=0\Rightarrow t=0\\x=ln2\Rightarrow t=1\)
Ta có:
\(\begin{aligned} & \int\limits_{0}^{\ln 2}{\sqrt{{{e}^{x}}-1}}dx=\int\limits_{0}^{1}{\frac{2{{t}^{2}}dt}{{{t}^{2}}+1}}=\int\limits_{0}^{1}{\left( 2-\frac{2}{{{t}^{2}}+1} \right)dt} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2t-2\arctan t\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2-2.\frac{\pi }{4} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2-\frac{\pi }{2} \\ \end{aligned} \)
c) Đặt \(t=\sqrt[3]{1-x}\Rightarrow t^3=1-x\Rightarrow dx=-3t^2dt \)
\(x=1\Rightarrow t=0\\x=9\Rightarrow t=-2\)