Giải bài 7 trang 145 SGK giải tích nâng cao 12
Tìm nguyên hàm của các hàm số sau:
a) \(f\left( x \right)=3x\sqrt{7-3{{x}^{2}}};\) b) \(f\left( x \right)=\cos \left( 3x+4 \right);\)
c) \(f\left( x \right)=\dfrac{1}{{{\cos }^{2}}\left( 3x+2 \right)};\) d) \(f\left( x \right)={{\sin }^{5}}\dfrac{x}{3}\cos \dfrac{x}{3}.\)
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
+ \(\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}+C\)
+ \(\int{\dfrac{1}{{{\cos }^{2}}\left( ax+b \right)}dx}=\dfrac{\tan \left( ax+b \right)}{a}+C\)