a) Đặt \(u = \sqrt {{t^5} + 2t} \Rightarrow {u^2} = {t^5} + 2t \Rightarrow 2udu = \left( {5{t^4} + 2} \right)dt\)
\(\int\limits_0^1 {\sqrt {{t^5} + 2t} } \left( {2 + 5{t^4}} \right)dt = \int\limits_0^{\sqrt 3 } {2{u^2}du = \left. {{{2{u^3}} \over 3}} \right|} _0^{\sqrt 3 } = 2\sqrt 3 \)
b) Ta có \(I = \int\limits_0^{{\pi \over 2}} {x\sin x\cos xdx = {1 \over 2}} \int\limits_0^{{\pi \over 2}} {x\sin 2xdx} \)
Đặt
\(\left\{ \matrix{ u = x \hfill \cr dv = \sin 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{ du = dx \hfill \cr v = - {1 \over 2}\cos 2x \hfill \cr} \right.\)
Do đó \(I = \left. {{1 \over 2}\left( { - {1 \over 2}x\cos x2x} \right)} \right|_0^{{\pi \over 2}} + {1 \over 4}\int\limits_0^{{\pi \over 2}} {\cos 2xdx = {\pi \over 8}} + \left. {{1 \over 8}\sin 2x} \right|_0^{{\pi \over 2}} = {\pi \over 8}\)
