Giải bài 9 trang 146 SGK giải tích nâng cao 12
Tìm nguyên hàm của các hàm số sau:
a) \(f\left( x \right)={{x}^{2}}\cos 2x;\) b) \(f\left( x \right)=\sqrt{x}\ln x;\)
c) \(f\left( x \right)={{\sin }^{4}}x\cos x;\) d) \(f\left( x \right)=x\cos \left( {{x}^{2}} \right).\)
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 \({{x}^{2}}=t\Leftrightarrow 2xdx=dt \)
\(\Rightarrow \int{x\cos \left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int{\cos tdt} \\ =\dfrac{1}{2}\sin t+C=\dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C \)