Giải bài 9 trang 146 SGK giải tích nâng cao 12

a) $\int{{{x}^{2}}\cos 2xdx}$

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

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

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

$\begin{aligned} \Rightarrow \int{x\sin 2xdx}&=-\dfrac{x\cos 2x}{2}+\int{\dfrac{\cos 2x}{2}dx} \\ & =-\dfrac{x\cos 2x}{2}+\dfrac{\sin 2x}{4} \\ \end{aligned}$

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

b) $\int{\sqrt{x}\ln xdx}$

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

$\begin{aligned} \Rightarrow \int{\sqrt{x}\ln xdx}&=\dfrac{2}{3}x\sqrt{x}\ln x-\dfrac{2}{3}\int{\sqrt{x}dx} \\ & =\dfrac{2}{3}x\sqrt{x}\ln x-\dfrac{4}{9}\sqrt{{{x}^{3}}}+C \\ \end{aligned}$

c) $\int{{{\sin }^{4}}x\cos xdx}$

Đặt $\sin x=t\Leftrightarrow dt=\cos xdx$

$\Rightarrow \int{{{\sin }^{4}}x\cos xdx}=\int{{{t}^{4}}dt}=\dfrac{1}{5}{{t}^{5}}+C \\ =\dfrac{1}{5}{{\sin }^{5}}x+C$

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

Đặt ${{x}^{2}}=t\Leftrightarrow 2xdx=dt$

$\Rightarrow \int{x\cos \left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int{\cos tdt} \\ =\dfrac{1}{2}\sin t+C=\dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C$