Giải bài 17 trang 161 SGK giải tích nâng cao 12
Dùng phương pháp đổi biến số tính các tích phân sau:
a) \(\int\limits_{0}^{1}{\sqrt{x+1}dx}\); b) \(\int\limits_{0}^{\frac{\pi }{4}}{\dfrac{\tan x}{{{\cos }^{2}}x}dx}\); c) \(\int\limits_{0}^{1}{{{t}^{3}}{{\left( 1+{{t}^{4}} \right)}^{3}}dt}\);
d) \(\int\limits_{0}^{1}{\dfrac{5x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}dx}\); e) \(\int\limits_{0}^{\sqrt{3}}{\dfrac{4x}{\sqrt{{{x}^{2}}+1}}}dx\); f) \(\int\limits_{0}^{\frac{\pi }{6}}{\left( 1-\cos 3x \right)\sin 3xdx} \).
a) Đặt \(x+1=t\Rightarrow dx=dt\)
| x | 0 | 1 |
| t | 1 | 2 |
\(\Rightarrow \int\limits_{0}^{1}{\sqrt{x+1}dx}=\int\limits_{1}^{2}{\sqrt{t}dt} \\ =\dfrac{2}{3}\sqrt{{{t}^{3}}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}}=\dfrac{2}{3}\left( 2\sqrt{2}-1 \right) \right. \)
b) Đặt \(\tan x=u\Rightarrow \dfrac{1}{{{\cos }^{2}}x}dx=du \)
| x | 0 | \(\dfrac{\pi}{4}\) |
| u | 0 | 1 |
\(\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}{\dfrac{\tan x}{{{\cos }^{2}}x}dx}=\int\limits_{0}^{1}{tdt}=\dfrac{{{t}^{2}}}{2}\left| _{\begin{smallmatrix} \\ \\\\\\0 \end{smallmatrix}}^{\begin{smallmatrix} 1\\\\\\ \\ \end{smallmatrix}}=\dfrac{1}{2} \right. \\ \)
c) Đặt \(1+{{t}^{4}}=u\Rightarrow 4{{t}^{3}}dt=du\Rightarrow {{t}^{3}}dt=\dfrac{1}{4}du \\ \)
| t | 0 | 1 |
| u | 1 | 2 |
\( \Rightarrow \int\limits_{0}^{1}{{{t}^{3}}{{\left( 1+{{t}^{4}} \right)}^{3}}dt}=\dfrac{1}{4}\int\limits_{1}^{2}{{{u}^{3}}du} \\ =\dfrac{1}{16}{{u}^{4}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right.=1-\dfrac{1}{16}=\dfrac{15}{16} \)
d) Đặt \({{x}^{2}}+4=t\Rightarrow 2xdx=dt\Rightarrow xdx=\dfrac{1}{2}dt\)
| x | 0 | 1 |
| t | 4 | 5 |
\(\Rightarrow \int\limits_{0}^{1}{\dfrac{5x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}dx}=\dfrac{5}{2}\int\limits_{4}^{5}{\dfrac{dt}{{{t}^{2}}}} \\ =-\dfrac{5}{2}.\dfrac{1}{t}\left| _{\begin{smallmatrix} \\ 4 \end{smallmatrix}}^{\begin{smallmatrix} 5 \\ \end{smallmatrix}} \right.=-\dfrac{1}{2}+\dfrac{5}{8}=\dfrac{1}{8} \)
e) \({{x}^{2}}+1=t\Rightarrow 2xdx=dt\Rightarrow xdx=\dfrac{1}{2}dt\)
| x | 0 | \(\sqrt{3}\) |
| t | 1 | 4 |
\(\Rightarrow \int\limits_{0}^{\sqrt{3}}{\dfrac{4x}{\sqrt{{{x}^{2}}+1}}}dx=\int\limits_{1}^{4}{\dfrac{2dt}{\sqrt{t}}} \\ =4\sqrt{t}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 4 \\ \end{smallmatrix}} \right.=8-4=4 \)
f) Đặt \(1-\cos 3x=t\Rightarrow 3\sin 3xdx=dt \)
| x | 0 | \(\dfrac{\pi}{6}\) |
| t | 0 | 1 |
\(\Rightarrow \int\limits_{0}^{\frac{\pi }{6}}{\left( 1-\cos 3x \right)\sin 3xdx=\dfrac{1}{3}\int\limits_{0}^{1}{tdt}} \\ =\dfrac{1}{6}{{t}^{2}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.=\dfrac{1}{6} \)