Giải bài 7 trang 145 SGK giải tích nâng cao 12

a) $\int{3x\sqrt{7-3{{x}^{2}}}dx}$

Đặt $t=7-3{{x}^{2}}\Leftrightarrow dt=-6xdx\Leftrightarrow 3xdx=-\dfrac{dt}{2}$

$\begin{aligned} \Rightarrow \int{3x\sqrt{7-3{{x}^{2}}}dx}&=-\dfrac{1}{2}\int{\sqrt{t}dt}=-\dfrac{1}{3}\sqrt{{{t}^{3}}}+C \\ & =-\dfrac{1}{3}\sqrt{7-3{{x}^{2}}}+C \\ \end{aligned}$

b) $\int{\cos \left( 3x+4 \right)dx}$

Đặt $t=3x+4\Leftrightarrow dt=3dx\Leftrightarrow dx=\dfrac{dt}{3}$

$\begin{aligned} \Rightarrow \int{\cos \left( 3x+4 \right)dx}&=\dfrac{1}{3}\int{\cos tdt}=\dfrac{\sin t}{3}+C \\ & =\dfrac{\sin \left( 3x+4 \right)}{3}+C \\ \end{aligned}$

c) $\int{\dfrac{1}{{{\cos }^{2}}\left( 3x+2 \right)}dx}$

Đặt $t=3x+2\Leftrightarrow dt=3dx\Leftrightarrow dx=\dfrac{dt}{3}$

$\begin{aligned} \Rightarrow \int{\dfrac{1}{{{\cos }^{2}}\left( 3x+2 \right)}dx}&=\dfrac{1}{3}\int{\dfrac{dt}{{{\cos }^{2}}t}}=\dfrac{\tan t}{3}+C \\ & =\dfrac{\tan \left( 3x+2 \right)}{3}+C \\ \end{aligned}$

d) $\int{{{\sin }^{5}}\dfrac{x}{3}\cos \dfrac{x}{3}dx}$

Đặt $t=\sin \dfrac{x}{3}\Leftrightarrow dt=\dfrac{1}{3}\cos \dfrac{x}{3}dx\Leftrightarrow 3dt=\cos \dfrac{x}{3}dx$

$\begin{aligned} \Rightarrow \int{{{\sin }^{5}}\dfrac{x}{3}\cos \dfrac{x}{3}dx}&=3\int{{{t}^{5}}dt}=\dfrac{{{t}^{6}}}{2}+C \\ & =\dfrac{1}{2}{{\sin }^{6}}\dfrac{x}{3}+C \\ \end{aligned}$

Tổng quát

+ $\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}+C$

+ $\int{\dfrac{1}{{{\cos }^{2}}\left( ax+b \right)}dx}=\dfrac{\tan \left( ax+b \right)}{a}+C$