\(\begin{array}{l}a) \,\,\,\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\sqrt[3]{{{{\left( {1 - x} \right)}^2}}}dx} = \int\limits_{ - \frac{1}{2}}^{\,\frac{1}{2}} {{{\left( {1 - x} \right)}^{\frac{2}{3}}}dx} \\= \left. { - 1.\frac{{{{\left( {1 - x} \right)}^{\frac{5}{3}}}}}{{\frac{5}{3}}}} \right|_{ - \frac{1}{2}}^{\frac{1}{2}} = - \frac{3}{5}.\left[ {{{\left( {\frac{1}{2}} \right)}^{\frac{5}{3}}} - {{\left( {\frac{3}{2}} \right)}^{\frac{5}{3}}}} \right]\\= - \frac{3}{5}\left[ {\frac{1}{{\sqrt[3]{{{2^5}}}}} - \frac{{\sqrt[3]{{{3^5}}}}}{{\sqrt[3]{{{2^5}}}}}} \right] = - \frac{3}{5}\left[ {\frac{1}{{\sqrt[3]{{{2^3}{{.2}^2}}}}} - \frac{{\sqrt[3]{{{3^3}{{.3}^2}}}}}{{\sqrt[3]{{{2^3}{{.2}^2}}}}}} \right]\\= - \frac{3}{5}\left[ {\frac{1}{{2\sqrt[3]{4}}} - \frac{{3\sqrt[3]{9}}}{{2\sqrt[3]{4}}}} \right] = \frac{3}{{10\sqrt[3]{4}}}\left( {3\sqrt[3]{9} - 1} \right)\end{array}\)
\(b)\,\,\int\limits_0^{\frac{\pi }{2}} {\sin \left( {\frac{\pi }{4} - x} \right)dx} = \left. {\cos \left( {\frac{\pi }{4} - x} \right)} \right|_0^{\frac{\pi }{2}} = \cos \left( { - \frac{\pi }{4}} \right) - \cos \frac{\pi }{4} = 0\)
c) Ta có: \(\frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{x} - \frac{1}{{x + 1}}\)
\(\begin{array}{l}\Rightarrow \int\limits_{\frac{1}{2}}^2 {\frac{1}{{x\left( {x + 1} \right)}}dx} = \int\limits_{\frac{1}{2}}^2 {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx} \\= \left. {\left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_{\frac{1}{2}}^2 = \left. {\ln \left| {\frac{x}{{x + 1}}} \right|} \right|_{\frac{1}{2}}^2\\= \ln \frac{2}{3} - \ln \frac{1}{3} = \ln \left( {\frac{2}{3}:\frac{1}{3}} \right) = \ln 2\end{array}\).
\(\begin{array}{l}d)\,\,x{\left( {x + 1} \right)^2} = x\left( {{x^2} + 2x + 1} \right) = {x^3} + 2{x^2} + x\\\Rightarrow \int\limits_0^2 {x{{\left( {x + 1} \right)}^2}dx} = \int\limits_0^2 {\left( {{x^3} + 2{x^2} + x} \right)dx} \\= \left. {\left( {\frac{{{x^4}}}{4} + 2\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right)} \right|_0^2 = \frac{{34}}{3}\end{array}\)
\(\begin{array}{l}e)\,\,\frac{{1 - 3x}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{ - 3\left( {x + 1} \right) + 4}}{{{{\left( {x + 1} \right)}^2}}} = - \frac{3}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}}\\\Rightarrow \int\limits_{\frac{1}{2}}^2 {\frac{{1 - 3x}}{{{{\left( {x + 1} \right)}^2}}}dx} = \int\limits_{\frac{1}{2}}^2 {\left( { - \frac{3}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} \\= - 3\int\limits_{\frac{1}{2}}^2 {\frac{{dx}}{{x + 1}}} + 4\int\limits_{\frac{1}{2}}^2 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}} \\= - \left. {3\ln \left| {x + 1} \right|} \right|_{\frac{1}{2}}^2 - \left. {\frac{4}{{x + 1}}} \right|_{\frac{1}{2}}^2\\= - 3\left( {\ln 3 - \ln \frac{3}{2}} \right) - 4\left( {\frac{1}{3} - \frac{2}{3}} \right)\\= - 3\ln 2 + \frac{4}{3}\end{array}\)
g) Cách 1:
Đặt \(f(x) = sin3xcos5x\) ta có: \(f\left( { - x} \right) = \sin \left( { - 3x} \right)\cos \left( { - 5x} \right) = - \sin 3x\cos 5x = - f\left( x \right) \Rightarrow \) là hàm số lẻ, từ đó ta có: \(\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin 3x\cos 5xdx} = 0\).
Cách 2:
\(\begin{array}{l}\sin 3x\cos 5x = \frac{1}{2}\left( {\sin 8x - \sin 2x} \right)\\\Rightarrow \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin 3x\cos 5xdx} = \frac{1}{2}\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {\sin 8x - \sin 2x} \right)dx} \\= \frac{1}{2}\left. {\left( { - \frac{{\cos 8x}}{8} + \frac{{\cos 2x}}{2}} \right)} \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}\\= \frac{1}{2}\left( { - \frac{5}{8} - \left( { - \frac{5}{8}} \right)} \right) = 0\end{array}\)
