Bài tập 3.45 trang 181 SBT Toán 12

a) Ta có: \({\cos ^2}x = \frac{{1 + \cos 2x}}{2}\)

\( \Rightarrow \cos 2x.{\cos ^2}x = \frac{1}{2}\cos 2x\left( {1 + \cos 2x} \right)\)

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

Suy ra

\(\begin{array}{l}
\int \limits_0^{\frac{\pi }{4}} \cos 2x.{\cos ^2}xdx\\
 = \int \limits_0^{\frac{\pi }{4}} \left( {\frac{1}{2}\cos 2x + \frac{1}{4}\cos 4x + \frac{1}{4}} \right)dx\\
 = \left. {\left( {\frac{1}{4}\sin 2x + \frac{1}{{16}}\sin 4x + \frac{1}{4}x} \right)} \right|_0^{\frac{\pi }{4}}\\
 = \frac{1}{4} + \frac{\pi }{{16}}
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
\frac{{{e^x}}}{{{e^{2x}} - 1}} = \frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 1} \right)}}\\
 = \frac{1}{2}\left( {\frac{{{e^x}}}{{{e^x} - 1}} - \frac{{{e^x}}}{{{e^x} + 1}}} \right)
\end{array}\)

Khi đó 

\(\begin{array}{l}
\int \limits_{\frac{1}{2}}^1 \frac{{{e^x}}}{{{e^{2x}} - 1}}dx = \frac{1}{2}\int \limits_{\frac{1}{2}}^1 \left( {\frac{{{e^x}}}{{{e^x} - 1}} - \frac{{{e^x}}}{{{e^x} + 1}}} \right)dx\\
 = \frac{1}{2}\left[ {\int \limits_{\frac{1}{2}}^1 \frac{{{e^x}}}{{{e^x} - 1}}dx - \int \limits_{\frac{1}{2}}^1 \frac{{{e^x}}}{{{e^x} + 1}}dx} \right]\\
 = \frac{1}{2}\left[ {\int \limits_{\frac{1}{2}}^1 \frac{{d\left( {{e^x}} \right)}}{{{e^x} - 1}} - \int \limits_{\frac{1}{2}}^1 \frac{{d\left( {{e^x}} \right)}}{{{e^x} + 1}}} \right]\\
 = \frac{1}{2}\left. {\left[ {\ln \left| {{e^x} - 1} \right| - \ln \left| {{e^x} + 1} \right|} \right]_{\frac{1}{2}}^1} \right|\\
 = \frac{1}{2}\left. {\left[ {\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 1}}} \right|} \right]} \right|_{\frac{1}{2}}^1\\
 = \frac{1}{2}\left( {\ln \frac{{e - 1}}{{e + 1}} - \ln \frac{{\sqrt e  - 1}}{{\sqrt e  + 1}}} \right)\\
 = \frac{1}{2}\ln \frac{{\left( {e - 1} \right)\left( {\sqrt e  + 1} \right)}}{{\left( {e + 1} \right)\left( {\sqrt e  - 1} \right)}}
\end{array}\)

c) Ta có: 

\(\begin{array}{l}
\frac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{x + 1}}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}\\
 = \frac{1}{{x + 1}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}
\end{array}\)

Khi đó

\(\begin{array}{l}
\int \limits_0^1 \frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx\\
 = \int \limits_0^1 \frac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx + \int \limits_0^1 \frac{{\ln \left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx\\
 = I + J
\end{array}\)

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

Tính \(J = \int \limits_0^1 \frac{{\ln \left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx\)

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

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

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

Vậy:

\(\begin{array}{l}
\int \limits_0^1 \frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx\\
 = \frac{{{{\ln }^2}2 - \ln 2 + 1}}{2}
\end{array}\)

d) Ta có:

\(\begin{array}{l}
\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}\\
 = \frac{{\left( {x\sin x + \cos x} \right) + x\cos x}}{{x\sin x + \cos x}}\\
 = 1 + \frac{{x\cos x}}{{x\sin x + \cos x}}
\end{array}\)

Khi đó 

\(\begin{array}{l}
\int \limits_0^{\frac{\pi }{4}} \frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx\\
 = \int \limits_0^{\frac{\pi }{4}} \left( {1 + \frac{{x\cos x}}{{x\sin x + \cos x}}} \right)dx\\
 = \int \limits_0^{\frac{\pi }{4}} dx + \int \limits_0^{\frac{\pi }{4}} \frac{{x\cos x}}{{x\sin x + \cos x}}dx\\
 = \frac{\pi }{4} + I
\end{array}\)

với \(I = \int \limits_0^{\frac{\pi }{4}} \frac{{x\cos x}}{{x\sin x + \cos x}}dx\)

Đặt \(x\sin x + \cos x = u\)

\(\begin{array}{l}
 \Rightarrow du = \left( {\sin x + x\cos x - \sin x} \right)dx\\
 = x\cos xdx
\end{array}\)

\(\begin{array}{l}
 \Rightarrow I = \int \limits_1^{\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)} \frac{{du}}{u} = \left. {\ln \left| u \right|} \right|_1^{\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)}\\
 = \ln \left[ {\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)} \right]\\
 = \ln \frac{{\sqrt 2 }}{2} + \ln \left( {\frac{\pi }{4} + 1} \right)\\
 = \ln \left( {1 + \frac{\pi }{4}} \right) - \frac{1}{2}\ln 2
\end{array}\)

Vậy

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