Bài tập 3.18 trang 171 SBT Toán 12

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

Ta có: 

\(\begin{array}{l}
\int\limits_0^{\frac{\pi }{2}} x \cos 2xdx = \left. {\frac{1}{2}x\sin 2x} \right|_0^{\frac{\pi }{2}} - \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} \\
 = 0 + \left. {\frac{1}{4}\cos 2x} \right|_0^{\frac{\pi }{2}} = \frac{1}{4}\left( { - 1 - 1} \right) =  - \frac{1}{2}
\end{array}\)

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

\(\begin{array}{l}
\int\limits_0^{\ln 2} x {e^{ - 2x}}dx =  - \frac{1}{2}\left. {x{e^{ - 2x}}} \right|_0^{\ln 2} + \frac{1}{2}\int\limits_0^{\ln 2} {{e^{ - 2x}}dx} \\
 =  - \frac{1}{2}\left( {\ln 2.\frac{1}{4} - 0} \right) - \frac{1}{4}\left. {{e^{ - 2x}}} \right|_0^{\ln 2}\\
 =  - \frac{1}{2}\left( {\ln 2.\frac{1}{4}} \right) - \frac{1}{4}\left( {\frac{1}{4} - 1} \right)\\
 = \frac{3}{{16}} - \frac{1}{8}\ln 2
\end{array}\)

c) Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {2x + 1} \right)\\
dv = dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{2dx}}{{2x + 1}}\\
v = x
\end{array} \right.\)

ta có: 

\(\begin{array}{l}
\int\limits_0^1 {\ln } (2x + 1)dx = x\left. {\ln \left( {2x + 1} \right)} \right|_0^1 - \int\limits_0^1 {\frac{{2x}}{{2x + 1}}dx} \\
 = \ln 3 - \int\limits_0^1 {\left( {1 - \frac{1}{{2x + 1}}} \right)dx} \\
 = \ln 3 - \left. {\left[ {x - \frac{1}{2}\ln \left( {2x + 1} \right)} \right]} \right|_0^1\\
 = \ln 3 - \left( {1 - \frac{1}{2}\ln 3} \right)\\
 = \frac{3}{2}\ln 3 - 1
\end{array}\)

d) Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {x - 1} \right) - \ln \left( {x + 1} \right)\\
dv = dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right)dx\\
v = x
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
\int\limits_2^3 {\left[ {\ln \left( {x - 1} \right)} \right] - \ln \left( {x + 1} \right)dx} \\
 = \left. {\left[ {x\ln \left( {x - 1} \right) - x\ln \left( {x + 1} \right)} \right]} \right|_2^3 - \int\limits_2^3 {\left( {\frac{x}{{x - 1}} - \frac{x}{{x + 1}}} \right)dx} \\
 = 2\ln 3 - 3\ln 2 - \int\limits_2^3 {\left( {\frac{1}{{x - 1}} + \frac{1}{{x + 1}}} \right)dx} \\
 = 2\ln 3 - 3\ln 2 - \left. {\left[ {\ln \left( {x - 1} \right) + \ln \left( {x + 1} \right)} \right]} \right|_2^3\\
 = 2\ln 3 - 3\ln 2 - \left( {3\ln 2 - \ln 3} \right)\\
 = 3\ln 3 - 6\ln 2
\end{array}\)

e) \(I = \int \limits_{\frac{1}{2}}^2 \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx = \int \limits_{\frac{1}{2}}^2 {e^{x + \frac{1}{x}}}dx + \int \limits_{\frac{1}{2}}^2 \left( {x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}dx = {I_1} + {I_2}\)

Tính tích phân I₁

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

Ta có: 

\(\begin{array}{l}
{I_1} = \left. {x{e^{x + \frac{1}{x}}}} \right|_{\frac{1}{2}}^2 - \int\limits_{\frac{1}{2}}^2 {x\left( {1 - \frac{1}{{{x^2}}}} \right){e^{x + \frac{1}{x}}}dx} \\
 = 2{e^{\frac{5}{2}}} - \frac{1}{2}{e^{\frac{5}{2}}} - {I_2}\\
 = \frac{3}{2}{e^{\frac{5}{2}}} - {I_2}
\end{array}\)

Suy ra:

\(I = \frac{3}{2}{e^{\frac{5}{2}}}\)

g) Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos x{\sin ^2}xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{3}{\sin ^3}x
\end{array} \right.\)

Ta có: 

\(\begin{array}{l}
\int\limits_0^{\frac{\pi }{2}} x \cos x{\sin ^2}xdx = \frac{1}{3}\left. {x{{\sin }^3}x} \right|_0^{\frac{\pi }{2}} - \frac{1}{3}\int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}xdx} \\
 = \frac{\pi }{6} - \frac{1}{3}J
\end{array}\)

trong đó 

\(\begin{array}{l}
J = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}xdx}  = \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{3}{4}\sin x - \frac{1}{4}\sin 3x} \right)dx} \\
 = \left. {\left( { - \frac{3}{4}\cos x + \frac{1}{{12}}\cos 3x} \right)} \right|_0^{\frac{\pi }{2}}\\
 = \frac{3}{4} - \frac{1}{{12}} = \frac{2}{3}
\end{array}\)

Vậy \(\int \limits_0^{\frac{\pi }{2}} x\cos x{\sin ^2}xdx = \frac{\pi }{6} - \frac{2}{9}\)

-- Mod Toán 12 HỌC247