Bài 5. Tính:
a) \(\int_0^3 } dx\)
b) \(\int_1^{64} {} dx\)
c) \(\int_0^2 {{x^2}} {e^{3x}}dx\)
d) \(\int_0^\pi  {\sqrt {1 + \sin 2x} } dx\)
Trả lời:
a) Đặt \(t = \sqrt {1 + x} \) , ta được: \(x = t^2- 1, dx = 2t dt\)
Khi \(x = 0\) thì \(t = 1\), khi \(x = 3\) thì \(t = 2\)
Do đó: 
\( \int_0^3 } dx = \int_1^2 {{{{t^2} - 1} \over t}} .2tdt = 2\int_1^2 {({t^2} - 1)dt}\)
\(= 2({{{t^3}} \over 3} - t)\left| {_1^2} \right. = 2({8 \over 3} - 2 - {1 \over 3} + 1) = {8 \over 3} \)
b)
Ta có:
\(\int_1^{64} {} dx = \int_1^{64} {}} \over {{x^}}}} dx = \int_1^{64} {({x^{{-1 \over 3}}} + {x^})dx}\)
\(=({3 \over 2}{x^} + {6 \over 7}{x^})\left| {_1^{64}} \right. = {{1839} \over {14}} \)                                                                                                        
c) Ta có:
\( \int_0^2 {{x^2}} {e^{3x}}dx = {1 \over 3}\int_0^2 {{x^2}} d{e^{3x}} = {1 \over 3}{x^2}{e^{3x}}\left| {_0^2} \right.\)
\(- {2 \over 3}\int_0^2 {x{e^{3x}}} dx \)\(= {4 \over 3}{e^6} - {2 \over 9}(x{e^{3x}})\left| {_0^2} \right. + {2 \over {27}}\int_0^2 {{e^{3x}}} d(3x) \)
\(= {4 \over 3}{e^6} - {4 \over 9}{e^6} + {2 \over {27}}{e^{3x}}\left| {_0^2} \right. = {2 \over {27}}(13{e^6} - 1) \)
d)
Ta có:
\( \sqrt {1 + \sin 2x} = \sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x{\mathop{\rm cosx}\nolimits} }\)
\(= |{\mathop{\rm s}\nolimits} {\rm{inx}} + {\mathop{\rm cosx}\nolimits} | \)\(= \sqrt 2 |\sin (x + {\pi \over 4})| \)
\(=\left\{ \matrix{
\sqrt 2 \sin (x + {\pi \over 4}),x \in \left[ {0,{{3\pi } \over 4}} \right] \hfill \cr
\sqrt 2 \sin (x + {\pi \over 4}),X \in \left[ {{{3\pi } \over 4},\pi } \right] \hfill \cr} \right.\) 
Do đó: 
\( \int_0^\pi {\sqrt {1 + \sin 2x} } dx = \sqrt 2 \int_0^{{{3\pi } \over 4}} {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4})\)\( - \sqrt 2 \int_{{{3\pi } \over 4}}^\pi {\sin (x + {\pi \over 4}} )d(x + {\pi \over 4}) \)  \(= - \sqrt 2 \cos (x + {\pi \over 4})\left| {_0^{{{3\pi } \over 4}}} \right. + \sqrt 2 (x + {\pi \over 4})\left| {_{{{3\pi } \over 4}}^\pi } \right. = 2\sqrt 2 \)