Giải bài 3.7 trang 164 - SBT Đại số và Giải tích lớp 12
Bằng cách biến đổi các hàm số lượng giác, hãy tính:
a) \(\int{\sin}^{4}dx\)
b) \(\int{\dfrac{1}{{{\sin }^{3}}x}dx}\)
c) \(\int{\sin ^3{x}\cos ^4{x}dx}\)
d) \(\int{\sin ^4{x}\cos ^4{x}dx}\)
a)Ta có:
\(\begin{align} & {{\sin }^{4}}x={{\left( \frac{1-\cos 2x}{2} \right)}^{2}}=\frac{1}{4}\left( 1-2\cos 2x+{{\cos }^{2}}2x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4}\left( 1-2\cos 2x+\frac{1+\cos 4x}{2} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4}\left( \frac{3}{2}-2\cos 2x+\frac{\cos 4x}{2} \right) \\ \end{align} \)
Suy ra:
\(\begin{align} & \int{{{\sin }^{4}}xdx=}\frac{1}{4}\int{\left( \frac{3}{2}-2\cos 2x+\frac{\cos 4x}{2} \right)}dx \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4}\int{\frac{3}{2}dx-\frac{1}{2}\int{\cos 2xdx+\frac{1}{8}\int{\cos 4xdx}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C \\ \end{align} \)
b) \(I=\int{\dfrac{1}{{{\sin }^{3}}x}dx=\int{\dfrac{\sin x}{{{\sin }^{4}}x}dx=\int{\dfrac{\sin x}{{{\left( 1-{{\cos }^{2}}x \right)}^{2}}}}}}dx\)
Đặt \(u=\cos x\Rightarrow du=-\sin xdx\). Suy ra: \(I=-\int \dfrac{du}{(1-u^2)^2}\)
Ta có:
\(\begin{align} & \frac{2}{1-{{u}^{2}}}=\frac{1}{1-u}+\frac{1}{1+u} \\ & \Rightarrow \frac{4}{(1-{{u}^{2}})^2}={{\left( \frac{1}{1-u}+\frac{1}{1+u} \right)}^{2}}=\frac{1}{{{\left( 1-u \right)}^{2}}}+\frac{2}{\left( 1-u \right)\left( 1+u \right)}+\frac{1}{{{\left( 1+u \right)}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{{{\left( 1-u \right)}^{2}}}+\frac{1}{{{\left( 1+u \right)}^{2}}}+\frac{1}{1-u}+\frac{1}{1+u} \\ \end{align} \)
Do đó:
\(\begin{align} & I=-\frac{1}{4}\int{\left[ \frac{1}{{{\left( 1-u \right)}^{2}}}+\frac{1}{{{\left( 1+u \right)}^{2}}}+\frac{1}{1-u}+\frac{1}{1+u} \right]}du \\ & =-\frac{1}{4}\int{\frac{du}{{{\left( u-1 \right)}^{2}}}}-\frac{1}{4}\int{\frac{du}{{{\left( 1+u \right)}^{2}}}}-\frac{1}{4}\int{\frac{du}{1-u}}-\frac{1}{4}\int{\frac{du}{1+u}} \\ & =\frac{1}{4}.\frac{1}{u-1}+\frac{1}{4}.\frac{1}{u+1}+\frac{1}{4}.\ln \left| 1-u \right|-\frac{1}{4}\ln \left| 1+u \right|+C \\ & =\frac{1}{4}.\frac{2u}{{{u}^{2}}-1}+\frac{1}{4}.\ln \left| \frac{1-u}{1+u} \right|+C \\ & =-\frac{\cos x}{2{{\sin }^{2}}x}+\frac{1}{4}.\ln \left| \frac{1-\cos x}{1+\cos x} \right|+C \\ & =\frac{-\cos x}{2{{\sin }^{2}}x}+\frac{1}{2}\ln \left| \tan \frac{x}{2} \right|+C \\ \end{align} \)
c) \(J=\int{\sin ^3{x}\cos ^4{x}dx}=\int (1-\cos^2x)\cos^4x \sin x dx\)
Đặt \(u=\cos x\Rightarrow du=-\sin{x}dx\)
Ta có:
\(\begin{align} & J=-\int{\left( 1-{{u}^{2}} \right){{u}^{4}}du=-\int{\left( {{u}^{4}}-{{u}^{6}} \right)du}} \\ & \,\,\,=-\frac{1}{5}{{u}^{5}}+\frac{1}{7}{{u}^{7}}+C \\ & \,\,\,=-\frac{1}{5}{{\cos }^{5}}x+\frac{1}{7}{{\cos }^{7}}x+C \\ \end{align} \)
d)
\(\begin{align} & {{\sin }^{4}}x{{\cos }^{4}}x={{\left( \sin x\cos x \right)}^{4}}={{\left( \frac{1}{2}\sin 2x \right)}^{4}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{16}{{\left( \frac{1-\cos 4x}{2} \right)}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{64}\left( 1-2\cos 4x+{{\cos }^{2}}4x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{64}\left( \frac{3}{2}-2\cos 4x+\frac{1}{2}\cos 8x \right) \\ \end{align} \)
Ta có:
\(\begin{align} &\int{\sin ^4{x}\cos ^4{x}dx}=\frac{1}{64}\int{\left( \frac{3}{2}-2\cos 4x+\frac{1}{2}\cos 8x \right)dx} \\ & \,\,\,=\frac{1}{64}\left( \frac{3}{2}x-\frac{1}{2}\sin 4x+\frac{1}{16}\sin 8x \right)+C \\ \end{align} \)