Bài tập 3.19 trang 171 SBT Toán 12

Bài tập 3.19 trang 171 SBT Toán 12

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

Ta có: 

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

b) Ta có: 

\(\begin{array}{l}
{\log _2}\left( {x + 1} \right) = \frac{{\ln \left( {x + 1} \right)}}{{\ln 2}}\\
 \Rightarrow \frac{{{x^2} + x + 1}}{{x + 1}}{\log _2}\left( {x + 1} \right) = \frac{1}{{\ln 2}}\left[ {x\ln \left( {x + 1} \right) + \frac{{\ln \left( {x + 1} \right)}}{{x + 1}}} \right]\\
 \Rightarrow I = \int\limits_0^1 {\frac{{{x^2} + x + 1}}{{x + 1}}{{\log }_2}\left( {x + 1} \right)dx = \frac{1}{{\ln 2}}\left( {{I_1} + {I_2}} \right)} 
\end{array}\)

+) \({I_1} = \int \limits_0^1 x\ln \left( {x + 1} \right)dx\)

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

Ta có:

\(\begin{array}{l}
{I_1} = \frac{1}{2}{x^2}\left. {\ln \left( {x + 1} \right)} \right|_0^1 - \frac{1}{2}\int\limits_0^1 {\frac{{{x^2}}}{{x + 1}}dx} \\
 = \frac{1}{2}\ln 2 - \frac{1}{2}\int\limits_0^1 {\left( {x - 1 + \frac{1}{{x + 1}}} \right)dx} \\
 = \frac{1}{2}\ln 2 - \frac{1}{2}\left. {\left[ {\frac{{{x^2}}}{2} - x + \ln \left( {x + 1} \right)} \right]} \right|_0^1\\
 = \frac{1}{2}\ln 2 - \frac{1}{2}\left[ {\frac{1}{2} - 1 + \ln 2} \right] = \frac{1}{4}
\end{array}\)

+ Tính I₂

 \(\begin{array}{l}
{I_2} = \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx = \int\limits_0^1 {\ln \left( {x + 1} \right)d\left[ {\ln \left( {x + 1} \right)} \right]} } \\
 = \frac{1}{2}\left. {{{\ln }^2}\left( {x + 1} \right)} \right|_0^1 = \frac{1}{2}{\ln ^2}2
\end{array}\)

Ta có: \(I = \frac{1}{{\ln 2}}\left( {\frac{1}{4} + \frac{1}{2}{{\ln }^2}2} \right) = \frac{1}{{4\ln 2}} + \frac{1}{2}\ln 2\)

c) Ta có: \(\frac{{{x^2} - 1}}{{{x^4} + 1}} = \frac{{1 - \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} = \frac{{1 - \frac{1}{{{x^2}}}}}{{\left( {x + \frac{1}{x}} \right) - 2}}\)

Đặt \(t = x + \frac{1}{x} \Rightarrow dt = \left( {1 - \frac{1}{{{x^2}}}} \right)dx\), ta có:

\(\int\limits_{\frac{1}{2}}^1 {\frac{{{x^2} - 1}}{{{x^4} + 1}}dx = \int\limits_{\frac{5}{2}}^2 {\frac{{dt}}{{{t^2} - 2}} = \frac{1}{{2\sqrt 2 }}\ln \left. {\left| {\frac{{t - \sqrt 2 }}{{t + \sqrt 2 }}} \right|} \right|_{\frac{5}{2}}^2 = \frac{1}{{2\sqrt 2 }}\ln \frac{{6 - \sqrt 2 }}{{6 + \sqrt 2 }}} } \)

d) 

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

-- Mod Toán 12