Giải bài 12 trang 147 – SGK môn Giải tích lớp 12
Tính các tích phân sau bằng phương pháp đổi biến số:
a) \(\int\limits_{0}^{\frac{\pi }{24}}{\tan \left( \dfrac{\pi }{3}-4x \right)dx}\) \(\left(\text{đặt} \,u=\cos \left( \dfrac{\pi }{3}-4x \right)\right)\);
b) \(\int\limits_{\frac{\sqrt{3}}{5}}^{\frac{3}{5}}{\dfrac{dx}{9+25{{x}^{2}}}}\)\(\left(\text{đặt} \,x=\dfrac{3}{5}\tan t\right)\);
c) \(\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}x{{\cos }^{4}}xdx}\) (đặt \(u=\cos x\));
d) \(\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\dfrac{\sqrt{1+\tan x}}{{{\cos }^{2}}x}dx}\) (đặt \(u=\sqrt{1+\tan x}\)).
Gợi ý: Sử dụng phương pháp đổi biến để tính tích phân.
a) Đặt \(u=\cos \left( \dfrac{\pi }{3}-4x \right)\Rightarrow du=4\sin \left( \dfrac{\pi }{3}-4x \right)dx\)
Đổi cận
| x | 0 | \(\dfrac{\pi}{24}\) |
| u | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{3}}{2}\) |
\(\begin{aligned} \int\limits_{0}^{\frac{\pi }{24}}{\tan \left( \dfrac{\pi }{3}-4x \right)dx}&=\int\limits_{0}^{\frac{\pi }{24}}{\dfrac{\sin \left( \dfrac{\pi }{3}-4x \right)}{\cos \left( \dfrac{\pi }{3}-4x \right)}dx} \\ & =\dfrac{1}{4}\int\limits_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}{\dfrac{du}{u}} \\ & =\dfrac{1}{4}\ln \left| u \right|\left| _{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \right. \\ & =\dfrac{1}{4}\left( \ln \dfrac{\sqrt{3}}{2}-\ln \dfrac{1}{2} \right) \\ & =\dfrac{1}{8}\ln 3 \\ \end{aligned} \)
b) Đặt \(x=\dfrac{3}{5}\tan t\Rightarrow dx=\dfrac{3}{5{{\cos }^{2}}t}dt \\ \)
Đổi cận
| x | \(\dfrac{\sqrt{3}}{5}\) | \(\dfrac{3}{5}\) |
| t | \(\dfrac{\pi}{6}\) | \(\dfrac{\pi}{4}\) |
\(\begin{aligned} \int\limits_{\frac{\sqrt{3}}{5}}^{\frac{3}{5}}{\dfrac{dx}{9+25{{x}^{2}}}}&=\dfrac{3}{5}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\dfrac{1}{9+9{{\tan }^{2}}t}}.\dfrac{dt}{{{\cos }^{2}}t} \\ & =\dfrac{3}{5}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\dfrac{{{\cos }^{2}}t}{9}}.\dfrac{dt}{{{\cos }^{2}}t} \\ & =\dfrac{1}{15}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}}{dt} \\ & =\dfrac{1}{15}t\left| _{\frac{\pi }{6}}^{\frac{\pi }{4}} \right. \\ & =\dfrac{1}{15}\left( \dfrac{\pi }{4}-\dfrac{\pi }{6} \right) \\ & =\dfrac{\pi }{180} \\ \end{aligned}\)
c) Đặt \(u=\cos x\Rightarrow du=-\sin xdx\)
Đổi cận
| x | 0 | \(\dfrac{\pi}{2}\) |
| u | 1 | 0 |
\(\begin{aligned} & \int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}x{{\cos }^{4}}xdx}=\int\limits_{0}^{\frac{\pi }{2}}{\left( 1-{{\cos }^{2}}x \right){{\cos }^{4}}x.\sin xdx} \\ & =\int\limits_{0}^{1}{\left( 1-{{u}^{2}} \right).{{u}^{4}}du} \\ & =\int\limits_{0}^{1}{\left( {{u}^{4}}-{{u}^{6}} \right)du} \\ & =\left( \dfrac{{{u}^{5}}}{5}-\dfrac{{{u}^{7}}}{7} \right)\left| _{\begin{smallmatrix} \\ \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \\ \\ \end{smallmatrix}} \right. \\ & =\dfrac{1}{5}-\dfrac{1}{7} \\ & =\dfrac{2}{35} \\ \end{aligned} \)
d) Đặt \(u=\sqrt{1+\tan x}\Rightarrow du=\dfrac{1}{2{{\cos }^{2}}x\sqrt{1+\tan x}}dx\)
Đổi cận
| x | \(-\dfrac{\pi}{4}\) | \(\dfrac{\pi}{4}\) |
| u | 0 | \(\sqrt{2}\) |
\(\begin{aligned} \int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\dfrac{\sqrt{1+\tan x}}{{{\cos }^{2}}x}dx}&=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\dfrac{1+\tan x}{{{\cos }^{2}}x\sqrt{1+\tan x}}dx} \\ & =2\int\limits_{0}^{\sqrt{2}}{{{u}^{2}}du} \\ & =\dfrac{2}{3}{{u}^{3}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \sqrt{2} \\ \end{smallmatrix}} \right. \\ & =\dfrac{4\sqrt{2}}{3} \\ \end{aligned} \)