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Bài 11. Tính các tích phân sau bằng phương pháp tính tích phân từng phần
a) \(\int_1^{{e^4}} {\sqrt x } \ln xdx\)
b) \(\int_{{\pi \over 6}}^{{\pi \over 2}} {{{xdx} \over {{{\sin }^2}x}}} \)
c) \(\int_0^\pi {(\pi – x)\sin {\rm{x}}dx} \)
d) \(\int_{ – 1}^0 {(2x + 3){e^{ – x}}} dx\)
a)
\(\eqalign{
& \int_1^{{e^4}} {\sqrt x } \ln xdx = {\int_1^{{e^4}} {\ln xd({2 \over 3}} x^{{3 \over 2}}}) \cr
& = {2 \over 3}{x^{{3 \over 2}}}\ln x\left| {_1^{{e^4}}} \right. \int\limits_1^{{e^4}} {{2 \over 3}} .{x^{{3 \over 2}}}.d{\mathop{\rm lnx}\nolimits} \cr
& = {8 \over 3}{e^6} – {2 \over 3}{x^{{1 \over 2}}}dx = {8 \over 3}{e^6} – {4 \over 9}{x^{{2 \over 3}}}\left| {_1^{{e^4}}} \right. = {{20} \over 9}{e^6} + {4 \over 9} \cr} \)
b) Ta có:
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\(\eqalign{
& \int_{{\pi \over 6}}^{{\pi \over 2}} {{{xdx} \over {{{\sin }^2}x}}} = \int\limits_{{\pi \over 6}}^{{\pi \over 2}} {xd( – \cot x) = – x\cot x\left| {_{{\pi \over 6}}^{{\pi \over 2}}} \right.} + \int\limits_{{\pi \over 6}}^{{\pi \over 2}} {\cot xdx} \cr
& = {{\pi \sqrt 3 } \over 6} + \int\limits_{{\pi \over 6}}^{{\pi \over 2}} {{{d\sin x} \over {{\mathop{\rm s}\nolimits} {\rm{inx}}}}} = {{\pi \sqrt 3 } \over 6} + \ln |sinx|\left| {_{{\pi \over 6}}^{{\pi \over 2}}} \right. = {{\pi \sqrt 3 } \over 6} + \ln 2 \cr} \)
c) Ta có:
\(\eqalign{
& \int_0^\pi {(\pi – x)\sin {\rm{x}}dx} = \int\limits_0^\pi {(\pi – x)d( – {\mathop{\rm cosx}\nolimits} )} \cr
& = – (\pi – x)cosx\left| {_0^\pi } \right. + \int\limits_0^\pi {{\mathop{\rm cosxd}\nolimits} (\pi – x) = \pi – s{\rm{inx}}\left| {_0^\pi } \right.} = \pi \cr} \)
d) Ta có:
\(\eqalign{
& \int_{ – 1}^0 {(2x + 3){e^{ – x}}} dx = \int\limits_{ – 1}^0 {(2x + 3)d( – {e^{ – x}}} ) \cr
& = (2x + 3){e^{ – x}}\left| {_0^{ – 1}} \right. + \int\limits_{ – 1}^e {{e^{ – x}}} .2dx = e – 3 + 2{e^{ – x}}\left| {_0^1} \right. = 3e – 5 \cr} \)
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