Giải bài 18 trang 161 SGK giải tích nâng cao 12

a) Đặt $\left\{ \begin{aligned} & \ln x=u \\ & {{x}^{5}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & \dfrac{dx}{x}=du \\ & v=\dfrac{{{x}^{6}}}{6} \\ \end{aligned} \right.$

$\begin{aligned} \Rightarrow \int\limits_{1}^{2}{{{x}^{5}}\ln x}dx &=\dfrac{{{x}^{6}}}{6}.\ln x\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right.-\dfrac{1}{6}\int\limits_{1}^{2}{{{x}^{5}}dx} \\ & =\dfrac{32}{3}\ln 2-\dfrac{1}{36}{{x}^{6}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right. \\ & =\dfrac{32}{3}\ln 2-\dfrac{7}{4} \\ \end{aligned}$

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

$\begin{aligned} \Rightarrow \int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}dx}&=\left( x+1 \right){{e}^{x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.-\int\limits_{0}^{1}{{{e}^{x}}dx} \\ & =2e-1-{{e}^{x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.=e \\ \end{aligned}$

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

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

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

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

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

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