a) Ta có:
\(\displaystyle \int_0^{{\pi \over 2}} {\cos 2xsi{n^2}} xdx = {1 \over 2}\int_0^{{\pi \over 2}} {\cos 2x(1 - \cos 2x)dx}\)\(\displaystyle = {1 \over 2}\int_0^{{\pi \over 2}} {\left[ {\cos 2x - {{1 + \cos 4x} \over 2}} \right]} dx\)
\(\displaystyle = {1 \over 4}\int_0^{{\pi \over 2}} {(2\cos 2x - \cos 4x - 1)dx} \)\(\displaystyle = {1 \over 4}\left[ {\sin 2x - {{\sin 4x} \over 4} - x} \right]_0^{{\pi \over 2}} \displaystyle = - {1 \over 4}.{\pi \over 2} = {{ - \pi } \over 8} \)
b) Ta có: Xét \({2^x}-{2^{ - x}} ≥ 0 ⇔ x ≥ 0\).
Ta tách thành tổng của hai tích phân:
\(\int_{ - 1}^1 {|{2^x}} - {2^{ - x}}|dx = - \int_{ - 1}^0 ( {2^x} - {2^{ - x}})dx \)\(+ \int_0^1 ( {2^x} - {2^{ - x}})dx\)
\(\displaystyle = - ({{{2^x}} \over {\ln 2}} + {{{2^{ - x}}} \over {\ln 2}})\left| {_{ - 1}^0} \right. + ({{{2^x}} \over {\ln 2}} + {{{2^{ - x}}} \over {\ln 2}})\left| {_0^1} \right. \) \(\displaystyle = {1 \over {\ln 2}} \)
c) Ta có:
\(\displaystyle \int_1^2 {{{(x + 1)(x + 2)(x + 3)} \over {{x^2}}}} dx \) \(\displaystyle = \int_1^2 {{{{x^3} + 6{x^2} + 11x + 6} \over {{x^2}}}dx} \)
\(\displaystyle = \int_1^2 {(x + 6 + {{11} \over x}} + {6 \over {{x^2}}})dx\)
\(\displaystyle = \left[ {{{{x^2}} \over 2} + 6x + 11\ln |x| - {6 \over x}} \right]\left| {_1^2} \right. \)
\( \displaystyle = (2 + 12 + 11\ln 2 - 3) - ({1 \over 2} + 6 - 6) \)
\(\displaystyle = {{21} \over 2} + 11\ln 2 \)
d) Ta có:
\(\begin{array}{l}\int\limits_0^2 {\dfrac{1}{{{x^2} - 2x - 3}}dx = \int\limits_0^2 {\dfrac{1}{{\left( {x + 1} \right)\left( {x - 3} \right)}}dx} } \\ = \dfrac{1}{4}\int\limits_0^2 {\left( {\dfrac{1}{{x - 3}} - \dfrac{1}{{x + 1}}} \right)dx} \\ = \left. {\dfrac{1}{4}\left[ {\ln \left| {x - 3} \right| - \ln \left| {x + 1} \right|} \right]} \right|_0^2\\ = \dfrac{1}{4}\left[ { - \ln 3 - \ln 3} \right] = - \dfrac{1}{2}\ln 3.\end{array}\)
e) Ta có:
\(\eqalign{
& \int_0^{{\pi \over 2}} {{{({\mathop{\rm s}\nolimits} {\rm{inx}} + {\mathop{\rm cosx}\nolimits} )}^2}dx} = \int_0^{{\pi \over 2}} {(1 + \sin 2x)dx} \cr
& = \left[ {x - {{\cos 2x} \over 2}} \right]\left| {_0^{{\pi \over 2}}} \right. = {\pi \over 2} + 1. \cr} \)
g) Ta có:
\(\eqalign{
& I = \int_0^\pi {{{(x + {\mathop{\rm s}\nolimits} {\rm{inx)}}}^2}} dx\int_0^\pi {({x^2}} + 2x\sin x + {\sin ^2}x)dx \cr
& = \left[ {{{{x^3}} \over 3}} \right]\left| {_0^\pi } \right. + 2\int_0^\pi {x\sin xdx + {1 \over 2}} \int_0^\pi {(1 - \cos 2x)dx}. \cr} \)
Tính :\(J = \int_0^\pi {x\sin xdx} \)
Đặt \(u = x ⇒ u’ = 1\) và \(v’ = sinx ⇒ v = -cos x\)
Suy ra:
\(J = \left[ { - x{\mathop{\rm cosx}\nolimits} } \right]\left| {_0^\pi } \right. + \int_0^\pi {{\mathop{\rm cosxdx}\nolimits} = \pi + \left[ {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right]} \left| {_0^\pi } \right. = \pi \)
Do đó:
\(\eqalign{
& I = {{{\pi ^3}} \over 3} + 2\pi + {1 \over 2}\left[ {x - {{\sin 2x} \over 2}} \right]\left| {_0^{{\pi }}} \right. \cr
& = {{{\pi ^3}} \over 3} + 2\pi + {\pi \over 2} = {{2{\pi ^3} + 15\pi } \over 6}. \cr} \)
