Bài 3.45 trang 181 SBT giải tích 12

Tính các tích phân sau:

a) \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\cos 2x.{{\cos }^2}xdx} \)    b) \(\displaystyle  \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^{2x}} - 1}}dx} \)

c) \(\displaystyle  \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)

d) \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)

a) \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\cos 2x} .{\cos ^2}xdx\)

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

\(\displaystyle   = \frac{1}{2}\cos 2x + \frac{1}{2}{\cos ^2}2x\) \(\displaystyle   = \frac{1}{2}\cos 2x + \frac{1}{4}\left( {1 + \cos 4x} \right)\) \(\displaystyle   = \frac{1}{2}\cos 2x + \frac{1}{4}\cos 4x + \frac{1}{4}\)

Suy ra \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\cos 2x} .{\cos ^2}xdx\)\(\displaystyle   = \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{2}\cos 2x + \frac{1}{4}\cos 4x + \frac{1}{4}} \right)dx} \) \(\displaystyle   = \left. {\left( {\frac{1}{4}\sin 2x + \frac{1}{{16}}\sin 4x + \frac{1}{4}x} \right)} \right|_0^{\frac{\pi }{4}}\) \(\displaystyle   = \frac{1}{4} + \frac{\pi }{{16}}\)

b) \(\displaystyle  \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^{2x}} - 1}}dx} \)

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

Khi đó \(\displaystyle  \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^{2x}} - 1}}dx} \) \(\displaystyle   = \frac{1}{2}\int\limits_{\frac{1}{2}}^1 {\left( {\frac{{{e^x}}}{{{e^x} - 1}} - \frac{{{e^x}}}{{{e^x} + 1}}} \right)dx} \) \(\displaystyle   = \frac{1}{2}\left[ {\int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^x} - 1}}dx}  - \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^x} + 1}}} dx} \right]\)

\(\displaystyle   = \frac{1}{2}\left[ {\int\limits_{\frac{1}{2}}^1 {\frac{{d\left( {{e^x}} \right)}}{{{e^x} - 1}}}  - \int\limits_{\frac{1}{2}}^1 {\frac{{d\left( {{e^x}} \right)}}{{{e^x} + 1}}} } \right]\) \(\displaystyle   = \frac{1}{2}\left. {\left[ {\ln \left| {{e^x} - 1} \right| - \ln \left| {{e^x} + 1} \right|} \right]} \right|_{\frac{1}{2}}^1\) \(\displaystyle   = \frac{1}{2}\left. {\left[ {\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 1}}} \right|} \right]} \right|_{\frac{1}{2}}^1\)

\(\displaystyle   = \frac{1}{2}\left( {\ln \frac{{e - 1}}{{e + 1}} - \ln \frac{{\sqrt e  - 1}}{{\sqrt e  + 1}}} \right)\) \(\displaystyle   = \frac{1}{2}\ln \frac{{\left( {e - 1} \right)\left( {\sqrt e  + 1} \right)}}{{\left( {e + 1} \right)\left( {\sqrt e  - 1} \right)}}\).

c) \(\displaystyle  \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)

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

Khi đó \(\displaystyle  \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)\(\displaystyle   = \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx}  + \int\limits_0^1 {\frac{{\ln \left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} \) \(\displaystyle   = I + J\)

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

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

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

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

Vậy \(\displaystyle  \int\limits_0^1 {\frac{{x + 2}}{{{x^2} + 2x + 1}}\ln (x + 1)dx} \)\(\displaystyle   = \frac{{{{\ln }^2}2}}{2} + \frac{1}{2} - \frac{{\ln 2}}{2}\) \(\displaystyle   = \frac{{{{\ln }^2}2 - \ln 2 + 1}}{2}\)

d) \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)

Ta có: \(\displaystyle  \frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}\)\(\displaystyle   = \frac{{\left( {x\sin x + \cos x} \right) + x\cos x}}{{x\sin x + \cos x}}\) \(\displaystyle   = 1 + \frac{{x\cos x}}{{x\sin x + \cos x}}\)

Khi đó \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)\(\displaystyle   = \int\limits_0^{\frac{\pi }{4}} {\left( {1 + \frac{{x\cos x}}{{x\sin x + \cos x}}} \right)dx} \) \(\displaystyle   = \int\limits_0^{\frac{\pi }{4}} {dx}  + \int\limits_0^{\frac{\pi }{4}} {\frac{{x\cos x}}{{x\sin x + \cos x}}dx} \)

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

Đặt \(\displaystyle  x\sin x + \cos x = u\) \(\displaystyle   \Rightarrow du = \left( {\sin x + x\cos x - \sin x} \right)dx\) \(\displaystyle   = x\cos xdx\)

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

Vậy \(\displaystyle  \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)\(\displaystyle   = \frac{\pi }{4} + \ln \left( {1 + \frac{\pi }{4}} \right) - \frac{1}{2}\ln 2\).