Giải bài 25 trang 162 SGK giải tích nâng cao 12

a) Đặt $\left\{ \begin{aligned} & x=u \\ & \cos 2xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & dx=du \\ & v=\dfrac{1}{2}\sin 2x \\ \end{aligned} \right.$

$\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}{x\cos 2xdx}=\dfrac{x}{2}\sin 2x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{4}} \right.-\dfrac{1}{2}\int\limits_{0}^{\frac{\pi }{4}}{\sin 2xdx} \\ =\dfrac{\pi }{8}+\dfrac{1}{4}\cos 2x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{4}} \right.=\dfrac{\pi }{8}-\dfrac{1}{4}$

b) Đặt $\ln \left( 2-x \right)=t\Rightarrow \dfrac{-dx}{2-x}=dt$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{\ln \left( 2-x \right)}{2-x}dx}=\int\limits_{0}^{\ln 2}{tdt}=\dfrac{{{t}^{2}}}{2}\left| _{\begin{smallmatrix} \\ \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \ln 2 \\ \\ \end{smallmatrix}} \right.=\dfrac{{{\left( \ln 2 \right)}^{2}}}{2}$

c) Đặt $\left\{ \begin{aligned} & {{x}^{2}}=u \\ & \cos xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & 2xdx=du \\ & v=\operatorname{sinx} \\ \end{aligned} \right.$

$\begin{align}\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{{{x}^{2}}\cos xdx}&={{x}^{2}}\sin x\left| _{\begin{smallmatrix} \\ \\\\ 0 \end{smallmatrix}}^{\frac{\pi }{2}}-2\int\limits_{0}^{\frac{\pi }{2}}{x\sin xdx} \right. \\ & =\dfrac{{{\pi }^{2}}}{4}-2\int\limits_{0}^{\frac{\pi }{2}}{x\sin xdx} \end{align}$

Đặt $\left\{ \begin{aligned} & x=u \\ & \sin xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & dx=du \\ & v=-\cos x \\ \end{aligned} \right.$

$\begin{align}\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{x\sin xdx}&=-x\cos x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.+\int\limits_{0}^{\frac{\pi }{2}}{\cos xdx} \\ & =\sin x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.=1\end{align}$

$\Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{{{x}^{2}}\cos xdx}=\dfrac{{{\pi }^{2}}}{4}-2$.

d) Đặt $\sqrt{{{x}^{3}}+1}=t\Rightarrow {{x}^{3}}+1={{t}^{2}}\Rightarrow 3{{x}^{2}}dx=2tdt$

$\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}\sqrt{{{x}^{3}}+1}dx}=\dfrac{2}{3}\int\limits_{1}^{\sqrt{2}}{t^2dt}=\dfrac{2}{9}{{t}^{3}}\left| _{\begin{smallmatrix} \\ \\\\ 1 \end{smallmatrix}}^{\begin{smallmatrix} \sqrt{2} \\ \\\\ \end{smallmatrix}} \right.=\dfrac{2}{9}(2\sqrt 2-1)$

e) Đặt $\left\{ \begin{aligned} & \ln x=u \\ & {{x}^{2}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & \dfrac{1}{x}dx=du \\ & v=\dfrac{{{x}^{3}}}{3} \\ \end{aligned} \right.$

$\Rightarrow \int\limits_{1}^{e}{{{x}^{2}}\ln xdx}=\dfrac{{{x}^{3}}}{3}\ln x\left| _{\begin{smallmatrix} \\ \\\\ 1 \end{smallmatrix}}^{\begin{smallmatrix} e \\ \\\\ \end{smallmatrix}} \right.-\dfrac{1}{3}\int\limits_{1}^{e}{{{x}^{2}}dx} \\ =\dfrac{{{e}^{3}}}{3}-\dfrac{1}{9}{{x}^{3}}\left| _{\begin{smallmatrix} \\ \\\\ 1 \end{smallmatrix}}^{\begin{smallmatrix} e \\ \\\\ \end{smallmatrix}} \right.=\dfrac{2{{e}^{3}}+1}{9}$