Giải bài 19 -20 trang 161 SGK giải tích nâng cao 12
Tính
19.a) \(\int\limits_{0}^{1}{\sqrt{{{t}^{5}}+2t}\left( 2+5{{t}^{4}} \right)dt}\); b) \(\int\limits_{0}^{\frac{\pi }{2}}{x\sin x\cos xdx}\).
20.a) \(\int\limits_{0}^{\pi }{5{{\left( 5-4\cos t \right)}^{\frac{1}{4}}}\sin tdt}\); b) \(\int\limits_{0}^{\sqrt{3}}{\dfrac{{{x}^{3}}dx}{\sqrt{{{x}^{2}}+1}}}\).
19.a) Đặt \({{t}^{5}}+2t=u\Rightarrow \left( 5{{t}^{4}}+2 \right)dt=du \)
| t | 0 | 1 |
| u | 0 | 3 |
\(\Rightarrow \int\limits_{0}^{1}{\sqrt{{{t}^{5}}+2t}\left( 2+5{{t}^{4}} \right)dt}=\int\limits_{0}^{3}{\sqrt{u}du} \\ =\dfrac{2}{3}\sqrt{{{u}^{3}}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 3 \\ \end{smallmatrix}} \right.=\dfrac{2}{3}.3\sqrt{3}=2\sqrt{3} \)
b) \(\int\limits_{0}^{\frac{\pi }{2}}{x\sin x\cos xdx}=\dfrac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{x\sin 2xdx}\)
Đặt \(\left\{ \begin{aligned} & x=u \\ & \sin 2xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & dx=du \\ & v=-\dfrac{1}{2}\cos 2x \\ \end{aligned} \right. \)
\(\begin{aligned} \Rightarrow \dfrac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{x\sin 2xdx}&=-\dfrac{1}{4}x.\cos 2x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.+\dfrac{1}{4}\int\limits_{0}^{\frac{\pi }{2}}{\cos 2xdx} \\ & =\dfrac{\pi }{8}+\dfrac{1}{8}\sin 2x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.=\dfrac{\pi }{8} \\ \end{aligned} \)
20.a) Đặt \(5-4\cos t=u\Rightarrow 4\sin tdt=du \)
| t | 0 | \(\pi\) |
| u | 1 | 9 |
\(\Rightarrow \int\limits_{0}^{\pi }{5{{\left( 5-4\cos t \right)}^{\frac{1}{4}}}\sin tdt}=\dfrac{5}{4}\int\limits_{1}^{9}{{{u}^{\frac{1}{4}}}du} \\ ={{u}^{\frac{5}{4}}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 9 \\ \end{smallmatrix}} \right.=9\sqrt{3}-1 \)
b) Đặt \(\sqrt{{{x}^{2}}+1}=t\Rightarrow {{x}^{2}}+1={{t}^{2}}\Rightarrow \left\{ \begin{aligned} & {{x}^{2}}={{t}^{2}}-1 \\ & 2xdx=2tdt \\ \end{aligned} \right. \)
| x | 0 | \(\sqrt 3\) |
| t | 1 | 2 |
\(\Rightarrow \int\limits_{0}^{\sqrt{3}}{\dfrac{{{x}^{3}}dx}{\sqrt{{{x}^{2}}+1}}}=\int\limits_{1}^{2}{\dfrac{{{t}^{2}}-1}{t}.tdt} \\ =\int\limits_{1}^{2}{\left( {{t}^{2}}-1 \right)dt}=\left( \dfrac{1}{3}{{t}^{3}}-t \right)\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right. \\ =\dfrac{8}{3}-2-\dfrac{1}{3}+1=\dfrac{4}{3} \)