Bài tập 3.7 trang 164 SBT Toán 12

Bài tập 3.7 trang 164 SBT Toán 12

a)Ta có:

\(\begin{array}{l}
{\sin ^4}x = {\left( {\frac{{1 - \cos 2x}}{2}} \right)^2} = \frac{1}{4}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)\\
 = \frac{1}{4}\left( {1 - 2\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)\\
 = \frac{1}{4}\left( {\frac{3}{2} - 2\cos 2x + \frac{{\cos 4x}}{2}} \right)
\end{array}\)

Suy ra: 

\(\begin{array}{l}
\int {{{\sin }^4}xdx}  = \frac{1}{4}\int {\left( {\frac{3}{2} - 2\cos 2x + \frac{{\cos 4x}}{2}} \right)} dx\\
 = \frac{1}{4}\int {\frac{3}{2}dx}  - \frac{1}{2}\int {\cos 2xdx}  + \frac{1}{8}\int {\cos 4xdx} \\
 = \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{{32}}\sin 4x + C
\end{array}\)

b) \(I = \int {\frac{1}{{{{\sin }^3}x}}dx}  = \int {\frac{{\sin x}}{{{{\sin }^4}x}}dx}  = \int {\frac{{\sin x}}{{{{\left( {1 - {{\cos }^2}x} \right)}^2}}}dx} \)

Đặt \(u = \cos x \Rightarrow du =  - \sin xdx\).

Suy ra: \(I =  - \int {\frac{{du}}{{{{(1 - {u^2})}^2}}}} \)

Ta có: 

\(\begin{array}{l}
\frac{2}{{1 - {u^2}}} = \frac{1}{{1 - u}} + \frac{1}{{1 + u}}\\
 \Rightarrow \frac{4}{{{{\left( {1 - {u^2}} \right)}^2}}} = {\left( {\frac{1}{{1 - u}} + \frac{1}{{1 + u}}} \right)^2}\\
 = \frac{1}{{{{\left( {1 - u} \right)}^2}}} + \frac{2}{{\left( {1 - u} \right)\left( {1 + u} \right)}} + \frac{1}{{{{\left( {1 + u} \right)}^2}}}\\
 = \frac{1}{{{{\left( {1 - u} \right)}^2}}} + \frac{1}{{{{\left( {1 + u} \right)}^2}}} + \frac{1}{{1 - u}} + \frac{1}{{1 + u}}
\end{array}\)

Do đó: 

\(\begin{array}{l}
I =  - \frac{1}{4}\smallint \left[ {\frac{1}{{{{\left( {1 - u} \right)}^2}}} + \frac{1}{{{{\left( {1 + u} \right)}^2}}} + \frac{1}{{1 - u}} + \frac{1}{{1 + u}}} \right]du\\
 =  - \frac{1}{4}\smallint \frac{{du}}{{{{\left( {u - 1} \right)}^2}}} - \frac{1}{4}\smallint \frac{{du}}{{{{\left( {1 + u} \right)}^2}}} - \frac{1}{4}\smallint \frac{{du}}{{1 - u}} - \frac{1}{4}\smallint \frac{{du}}{{1 + u}}\\
 = \frac{1}{4}.\frac{1}{{u - 1}} + \frac{1}{4}.\frac{1}{{u + 1}} + \frac{1}{4}.\ln \left| {1 - u} \right| - \frac{1}{4}\ln \left| {1 + u} \right| + C\\
 = \frac{1}{4}.\frac{{2u}}{{{u^2} - 1}} + \frac{1}{4}.\ln \left| {\frac{{1 - u}}{{1 + u}}} \right| + C\\
 =  - \frac{{\cos x}}{{2{{\sin }^2}x}} + \frac{1}{4}\ln \left| {\frac{{1 - \cos x}}{{1 + \cos x}}} \right| + C\\
 = \frac{{ - \cos x}}{{2{{\sin }^2}x}} + \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| + C
\end{array}\)

c) \(J = \int {{{\sin }^3}x{{\cos }^4}xdx}  = \int {(1 - {{\cos }^2}x){{\cos }^4}x\sin xdx} \)

Đặt \(u = \cos x \Rightarrow du =  - \sin xdx\)

Ta có: 

\(\begin{array}{l}
J =  - \mathop \smallint \nolimits \left( {1 - {u^2}} \right){u^4}du =  - \mathop \smallint \nolimits \left( {{u^4} - {u^6}} \right)du\\
 =  - \frac{1}{5}{u^5} + \frac{1}{7}{u^7} + C\\
 =  - \frac{1}{5}{\cos ^5}x + \frac{1}{7}{\cos ^7}x + C
\end{array}\)

d)

\(\begin{array}{l}
{\sin ^4}x{\cos ^4}x = {\left( {\sin x\cos x} \right)^4} = {\left( {\frac{1}{2}\sin 2x} \right)^4}\\
 = \frac{1}{{16}}{\left( {\frac{{1 - \cos 4x}}{2}} \right)^2} = \frac{1}{{64}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)\\
 = \frac{1}{{64}}\left( {\frac{3}{2} - 2\cos 4x + \frac{1}{2}\cos 8x} \right)
\end{array}\)

Ta có:

\(\begin{array}{l}
\int {{{\sin }^4}x{{\cos }^4}xdx}  = \frac{1}{{64}}\int {\left( {\frac{3}{2} - 2\cos 4x + \frac{1}{2}\cos 8x} \right)dx} \\
 = \frac{1}{{64}}\left( {\frac{3}{2}x - \frac{1}{2}\sin 4x + \frac{1}{{16}}\sin 8x} \right) + C
\end{array}\)

-- Mod Toán 12