Giải bài 6 trang 145 SGK giải tích nâng cao 12

a) $\int{x\sin \dfrac{x}{2}dx}$

Đặt $\left\{ \begin{aligned} & u=x \\ & \sin \dfrac{x}{2}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=dx \\ & v=-2\cos \dfrac{x}{2} \\ \end{aligned} \right.$

$\Rightarrow \int{x\sin \dfrac{x}{2}dx}=-2x\cos \dfrac{x}{2}+2\int{\cos \dfrac{x}{2}dx} \\ =-2x\cos \dfrac{x}{2}+4\sin \dfrac{x}{2}+C$

b) $\int{{{x}^{2}}\cos xdx}$

Đặt $\left\{ \begin{aligned} & u={{x}^{2}} \\ & \cos xdx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=2xdx \\ & v=\sin x \\ \end{aligned} \right.$

$\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\int{x\sin xdx}$

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

$\begin{aligned} \Rightarrow \int{x\sin xdx}&=-x\cos x+\int{\cos xdx} \\ & =-x\cos x+\sin x \\ \end{aligned}$

$\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\left( -x\cos x+\sin x \right) \\ ={{x}^{2}}\sin x+2x\cos x-2\sin x+C$

c) $\int{x{{e}^{x}}dx}$

Đặt $\left\{ \begin{aligned} & u=x \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=dx \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\begin{align}\Rightarrow \int{x{{e}^{x}}dx}&=x{{e}^{x}}-\int{{{e}^{x}}dx} \\ & =x{{e}^{x}}-{{e}^{x}}+C\end{align}$

d) $\int{{{x}^{3}}\ln \left( 2x \right)dx}$

Đặt $\left\{ \begin{aligned} & u=\ln \left( 2x \right) \\ & {{x}^{3}}dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=\dfrac{dx}{x} \\ & v=\dfrac{{{x}^{4}}}{4} \\ \end{aligned} \right.$

$\begin{aligned} \Rightarrow \int{{{x}^{3}}\ln \left( 2x \right)dx}&=\dfrac{{{x}^{4}}}{4}\ln \left( 2x \right)-\int{\dfrac{{{x}^{3}}}{4}dx} \\ & =\dfrac{{{x}^{4}}}{4}\ln \left( 2x \right)-\dfrac{{{x}^{4}}}{16}+C \\ \end{aligned}$