a) \(I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {x + 1} \right)\cos \left( {x + \dfrac{\pi }{2}} \right)} dx\)
Ta có: \(I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {x + 1} \right)\cos \left( {x + \dfrac{\pi }{2}} \right)} dx\) \( = - \int\limits_0^{\dfrac{\pi }{2}} {\left( {x + 1} \right)\sin x} dx\)
Đặt \(\left\{ \begin{array}{l}u = x + 1\\dv = \sin xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = - \cos x\end{array} \right.\)
\( \Rightarrow I = - \left[ { - \left. {\left( {x + 1} \right)\cos x} \right|_0^{\dfrac{\pi }{2}} + \int\limits_0^{\dfrac{\pi }{2}} {\cos xdx} } \right]\) \( = - \left( {1 + \left. {\sin x} \right|_0^{\dfrac{\pi }{2}}} \right) = - \left( {1 + 1} \right) = - 2\)
b) \(I = \int\limits_0^1 {\dfrac{{{x^2} + x + 1}}{{x + 1}}{{\log }_2}\left( {x + 1} \right)dx} \)
Ta có: \(\dfrac{{{x^2} + x + 1}}{{x + 1}}{\log _2}(x + 1)\)\( = \left( {x + \dfrac{1}{{x + 1}}} \right).\dfrac{{\ln \left( {x + 1} \right)}}{{\ln 2}}\) \( = \dfrac{1}{{\ln 2}}\left[ {x\ln (x + 1) + \dfrac{{\ln (x + 1)}}{{x + 1}}} \right]\)
Khi đó \(I = \int\limits_0^1 {\dfrac{{{x^2} + x + 1}}{{x + 1}}{{\log }_2}\left( {x + 1} \right)dx} \) \( = \dfrac{1}{{\ln 2}}\int\limits_0^1 {x\ln \left( {x + 1} \right)dx} \) \( + \dfrac{1}{{\ln 2}}\int\limits_0^1 {\dfrac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx} \)
Tính \(J = \int\limits_0^1 {x\ln \left( {x + 1} \right)dx} \).
Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = \dfrac{{{x^2}}}{2}\end{array} \right.\)
\( \Rightarrow J = \left. {\dfrac{{{x^2}}}{2}\ln \left( {x + 1} \right)} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 1}}dx} \) \( = \dfrac{{\ln 2}}{2} - \dfrac{1}{2}\int\limits_0^1 {\left( {x - 1 + \dfrac{1}{{x + 1}}} \right)dx} \) \( = \dfrac{1}{2}\ln 2 - \dfrac{1}{2}\left. {\left( {\dfrac{{{x^2}}}{2} - x + \ln \left( {x + 1} \right)} \right)} \right|_0^1\)
\( = \dfrac{1}{2}\ln 2 - \dfrac{1}{2}\left( {\dfrac{1}{2} - 1 + \ln 2} \right)\) \( = \dfrac{1}{4}\)
Tính \(K = \int\limits_0^1 {\dfrac{{\ln \left( {x + 1} \right)}}{{x + 1}}dx} \).
Đặt \(\ln \left( {x + 1} \right) = t \Rightarrow dt = \dfrac{{dx}}{{x + 1}}\) \( \Rightarrow K = \int\limits_0^{\ln 2} {tdt} = \left. {\dfrac{{{t^2}}}{2}} \right|_0^{\ln 2} = \dfrac{{{{\ln }^2}2}}{2}\)
Vậy \(I = \dfrac{1}{{\ln 2}}J + \dfrac{1}{{\ln 2}}K\) \( = \dfrac{1}{{4\ln 2}} + \dfrac{{\ln 2}}{2}\).
c) \(I = \int\limits_{\dfrac{1}{2}}^1 {\dfrac{{{x^2} - 1}}{{{x^4} + 1}}} dx\)
Đặt \(t = x + \dfrac{1}{x}\)\( \Rightarrow dt = 1 - \dfrac{1}{{{x^2}}}dx = \dfrac{{{x^2} - 1}}{{{x^2}}}dx\) và \({t^2} = {x^2} + 2 + \dfrac{1}{{{x^2}}} = \dfrac{{{x^4} + 1}}{{{x^2}}} + 2\) \( \Rightarrow \dfrac{{{x^2}}}{{{x^4} + 1}} = \dfrac{1}{{{t^2} - 2}}\).
Khi đó \(I = \int\limits_{\dfrac{1}{2}}^1 {\dfrac{{{x^2} - 1}}{{{x^4} + 1}}} dx\)\( = \int\limits_{\dfrac{1}{2}}^1 {\dfrac{{{x^2}}}{{{x^4} + 1}}.\dfrac{{{x^2} - 1}}{{{x^2}}}dx} \) \( = \int\limits_{\dfrac{5}{2}}^2 {\dfrac{{dt}}{{{t^2} - 2}}} \) \( = \dfrac{1}{{2\sqrt 2 }}\int\limits_{\dfrac{5}{2}}^2 {\left( {\dfrac{1}{{t - \sqrt 2 }} - \dfrac{1}{{t + \sqrt 2 }}} \right)dt} \)
\( = \left. {\ln \left| {\dfrac{{t - \sqrt 2 }}{{t + \sqrt 2 }}} \right|} \right|_{\dfrac{5}{2}}^2 = \dfrac{1}{{2\sqrt 2 }}\ln \dfrac{{6 - \sqrt 2 }}{{6 + \sqrt 2 }}\).
d) \(I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin 2xdx}}{{3 + 4\sin x - \cos 2x}}} \)
Ta có: \(\dfrac{{\sin 2x}}{{3 + 4\sin x - \cos 2x}}\) \( = \dfrac{{2\sin x\cos x}}{{3 + 4\sin x - 1 + 2{{\sin }^2}x}}\) \( = \dfrac{{\sin x\cos x}}{{{{\sin }^2} + 2\sin x + 1}}\) \( = \dfrac{{\sin x\cos x}}{{{{\left( {\sin x + 1} \right)}^2}}}\)
Khi đó \(I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x\cos x}}{{{{\left( {\sin x + 1} \right)}^2}}}dx} \).
Đặt \(\sin x = t \Rightarrow dt = \cos xdx\).
\( \Rightarrow I = \int\limits_0^1 {\dfrac{{tdt}}{{{{\left( {t + 1} \right)}^2}}}} \) \( = \int\limits_0^1 {\left( {\dfrac{1}{{t + 1}} - \dfrac{1}{{{{\left( {t + 1} \right)}^2}}}} \right)dt} \) \( = \left. {\left[ {\ln \left( {t + 1} \right) + \dfrac{1}{{t + 1}}} \right]} \right|_0^1\) \( = \ln 2 + \dfrac{1}{2} - 1 = \ln 2 - \dfrac{1}{2}\).
