Bài tập 3.6 trang 164 SBT Toán 12

Bài tập 3.6 trang 164 SBT Toán 12

a) \({I_1} = \mathop \smallint \nolimits x{\left( {3 - x} \right)^5}dx\)

Đặt \(3 - x = u \Rightarrow dx =  - du\), ta có:

\(\begin{array}{l}
{I_1} = \mathop \smallint \nolimits \left( {u - 3} \right){u^5}du = \mathop \smallint \nolimits \left( {{u^6} - 3{u^5}} \right)du = \frac{1}{7}{u^7} - \frac{1}{2}{u^6} + C\\
 = \frac{1}{7}{\left( {3 - x} \right)^7} - \frac{1}{2}{\left( {3 - x} \right)^6} + C
\end{array}\)

b) Ta có:

\(\begin{array}{l}
\smallint \left( {{2^x} - {3^x}} \right)2dx = \smallint \left( {{2^{2x}} - {{2.2}^x}{3^x} + {3^{2x}}} \right)dx\\
 = \smallint {4^x}dx - 2\smallint {6^x}dx + \smallint {9^x}dx\\
 = \frac{{{4^x}}}{{\ln 4}} - 2\frac{{{6^x}}}{{\ln 6}} + \frac{{{9^x}}}{{\ln 9}} + C
\end{array}\)

c) \({I_2} = \mathop \smallint \nolimits x\sqrt {2 - 5x} dx\)

Đặt \(\sqrt {2 - 5x}  = u \Rightarrow 2 - 5x = {u^2} \Rightarrow dx =  - \frac{2}{5}udu\)

Ta có: \(x = \frac{{2 - u}}{5}\) nên

\(\begin{array}{l}
{I_2} =  - \mathop \smallint \nolimits \frac{{2 - u}}{5}.u.\frac{2}{5}udu = \mathop \smallint \nolimits \left( { - \frac{4}{{25}}{u^2} + \frac{2}{{25}}{u^3}} \right)du\\
 =  - \frac{4}{{75}}{u^3} + \frac{1}{{50}}{u^4} + C\\
 =  - \frac{4}{{75}}{\left( {2 - 5x} \right)^{\frac{3}{2}}} + \frac{1}{{50}}{\left( {2 - 5x} \right)^2} + C
\end{array}\)

d) \({I_3} = \mathop \smallint \nolimits \frac{{\ln \left( {\cos x} \right)}}{{{{\cos }^2}x}}dx\)

Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {\cos x} \right)\\
dv = 1\cos 2xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du =  - \sin x\cos xdx\\
v = \tan x
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
{I_3} = \tan x\ln \left( {\cos x} \right) + \smallint \tan x.\frac{{\sin x}}{{\cos x}}dx\\
 = \tan x\ln \left( {\cos x} \right) + \smallint \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}dx\\
 = \tan x\ln \left( {\cos x} \right) + \smallint \frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}dx\\
 = \tan x\ln \left( {\cos x} \right) + \smallint \left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx\\
 = \tan x\ln \left( {\cos x} \right) + \tan x - x + C
\end{array}\)

e) \({I_4} = \mathop \smallint \nolimits \frac{x}{{{{\sin }^2}x}}dx\)

Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \frac{1}{{{{\sin }^2}x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v =  - \cot x
\end{array} \right.\)

Ta có: 

\({I_4} =  - x\cot x + {\rm{ }}\int {\cot xdx}  =  - x\cot x + {\rm{ }}\int {\frac{{\cos x}}{{\sin x}}} dx\)

Đặt \(t = \sin x \Rightarrow dt = \cos xdx\). Suy ra:

\({I_4} =  - x\cot x + \mathop \smallint \nolimits \frac{{dt}}{t} =  - x\cot x + \ln \left| t \right| + C =  - x\cot x + \ln \left| {{\rm{sinx}}} \right| + C\)

g) \({I_5} = \mathop \smallint \nolimits \frac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}dx\)

Ta có: \(\frac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 3} \right)}} = \frac{3}{{5(x - 2)}} + \frac{2}{{5(x + 3)}}\)

Khi đó:

\({{I_5} = \mathop \smallint \nolimits \frac{3}{{5\left( {x - 2} \right)}}dx + \mathop \smallint \nolimits \frac{2}{{5\left( {x + 3} \right)}}dx = \frac{3}{5}\ln \left| {x - 2} \right| + \frac{2}{5}\ln \left| {x + 3} \right| + C}\)

h) \({I_6} = \mathop \smallint \nolimits \frac{1}{{1 - \sqrt x }}dx\)

Đặt \(u = \sqrt x  \Rightarrow {u^2} = x \Rightarrow dx = 2udu\)

Ta có:

\(\begin{array}{l}
{I_6} = \mathop \smallint \nolimits \frac{{2udu}}{{1 - u}} = \mathop \smallint \nolimits \left( { - 2 + \frac{2}{{1 - u}}} \right)du\\
 =  - 2u + 2\ln \left| {1 - u} \right| + C\\
 =  - 2\sqrt x  + 2ln\left| {1 - \sqrt x } \right| + C
\end{array}\)

i) \(\int {\sin 3x} \cos 2xdx = \frac{1}{2}{\rm{ }}\int {(\sin x + \sin 5x)} dx =  - \frac{1}{2}\left( {\cos x + \frac{1}{5}\cos 5x} \right) + C\)

-- Mod Toán 12