Đáp án:

\(\begin{array}{l} h)\,\,\frac{\pi }{{16}}\\ k)\,\,\ln \left( {7 - 4\sqrt 3 } \right) \end{array}\)

Giải thích các bước giải: \[\begin{array}{l} h)\,\,I = \int\limits_0^1 {{x^2}\sqrt {1 - {x^2}} dx} \\ Dat\,x = \sin t \Rightarrow dx = {\mathop{\rm cost}\nolimits} dt\\ Doi\,\,can:\,\\ x = 0 \Rightarrow t = 0\\ x = 1 \Rightarrow t = \frac{\pi }{2}\\ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}t.\sqrt {1 - {{\sin }^2}t} \cos tdt} = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}t{{\cos }^2}tdt} = \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}2tdt} \\ = \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\frac{{1 - \cos 4t}}{2}dt} = \frac{1}{8}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 4t} \right)dt} = \frac{1}{8}\left. {\left( {t - \frac{{\sin 4t}}{4}} \right)} \right|_0^{\frac{\pi }{2}}\\ = \frac{1}{8}\left( {\frac{\pi }{2} - \frac{{\sin 2\pi }}{4} - 0 + \frac{{\sin 0}}{4}} \right) = \frac{1}{8}.\frac{\pi }{2} = \frac{\pi }{{16}}\\ k)\,\,I = \int\limits_0^1 {\sqrt {2x + {x^2}} dx} = \int\limits_0^1 {\sqrt {{{\left( {x + 1} \right)}^2} - 1} dx} \\ Dat\,\,x + 1 = \frac{1}{{\sin t}} \Rightarrow dx = - \frac{{\cos t}}{{{{\sin }^2}t}}dt\\ Doi\,\,can:\,\\ x = 0 \Rightarrow \sin t = 1 \Rightarrow t = \frac{\pi }{2}\\ x = 1 \Rightarrow \sin t = \frac{1}{2} \Rightarrow t = \frac{\pi }{6}\\ \Rightarrow I = - \int\limits_{\frac{\pi }{2}}^{\frac{\pi }{6}} {\sqrt {\frac{1}{{{{\sin }^2}t}} - 1} \frac{{\cos t}}{{{{\sin }^2}t}}dt} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\cos t}}{{\sin t}}.\frac{{\cos t}}{{{{\sin }^2}t}}dt} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{dt}}{{\sin t}}} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{dt\sin t}}{{1 - {{\cos }^2}t}}} \\ Dat\,\,u = \cos t \Rightarrow du = - \sin tdt\\ Doi\,\,can:\,\,t = \frac{\pi }{6} \Rightarrow u = \frac{{\sqrt 3 }}{2};\,\,t = \frac{\pi }{2} \Rightarrow u = 0\\ \Rightarrow I = - \int\limits_{\frac{{\sqrt 3 }}{2}}^0 {\frac{{du}}{{1 - {u^2}}}} = \int\limits_0^{\frac{{\sqrt 3 }}{2}} {\frac{{du}}{{\left( {1 - u} \right)\left( {1 + u} \right)}}} = \left. {\ln \left| {\frac{{1 - u}}{{1 + u}}} \right|} \right|_0^{\frac{{\sqrt 3 }}{2}}\\ = \ln \left( {7 - 4\sqrt 3 } \right) \end{array}\]