Bài 6 trang 127 SGK Giải tích 12

Đề bài

Tính:

a) \(\int_0^{{\pi  \over 2}} {\cos 2xsi{n^2}} xdx\)

b) \(\int_{ - 1}^1 {|{2^x}}  - {2^{ - x}}|dx\)

c) \(\int_1^2 {{{(x + 1)(x + 2)(x + 3)} \over {{x^2}}}} dx\)

d) \(\int_0^2 {{1 \over {{x^2} - 2x - 3}}} dx\)

e) \(\int_0^{{\pi  \over 2}} {{{({\mathop{\rm s}\nolimits} {\rm{inx}} + {\mathop{\rm cosx}\nolimits} )}^2}dx} \)

g) \(\int_0^\pi  {{{(x + {\mathop{\rm s}\nolimits} {\rm{inx}})}^2}} dx\)

Hướng dẫn giải

a) Ta có:

\( \int_0^{{\pi \over 2}} {\cos 2xsi{n^2}} xdx = {1 \over 2}\int_0^{{\pi \over 2}} {\cos 2x(1 - \cos 2x)dx}\)
\(= {1 \over 2}\int_0^{{\pi \over 2}} {\left[ {\cos 2x - {{1 + \cos 4x} \over 2}} \right]} dx\)

\( = {1 \over 4}\int_0^{{\pi \over 2}} {(2\cos 2x - \cos 4x - 1)dx} \)
\( = {1 \over 4}\left[ {\sin 2x - {{\sin 4x} \over 4} - x} \right]_0^{{\pi \over 2}} = - {1 \over 4}.{\pi \over 2} = {{ - \pi } \over 8} \)

b) Ta có: Xét \({2^x}-{2^{ - x}} ≥ 0 ⇔ x ≥ 0\).

Ta tách thành tổng của hai tích phân:

\(\int_{ - 1}^1 {|{2^x}} - {2^{ - x}}|dx = - \int_{ - 1}^0 ( {2^x} - {2^{ - x}})dx \)\(+ \int_0^1 ( {2^x} - {2^{ - x}})dx\)
\(= - ({{{2^x}} \over {\ln 2}} + {{{2^{ - x}}} \over {\ln 2}})\left| {_{ - 1}^0} \right. + ({{{2^x}} \over {\ln 2}} + {{{2^{ - x}}} \over {\ln 2}})\left| {_0^1} \right. \)\(= {1 \over {\ln 2}}  \)

c) Ta có: 

 \(\int_1^2 {{{(x + 1)(x + 2)(x + 3)} \over {{x^2}}}} dx = \int_1^2 {{{{x^3} + 6{x^2} + 11x + 6} \over {{x^2}}}dx} \) 
\(= \int_1^2 {(x + 6 + {{11} \over x}} + {6 \over {{x^2}}})dx\)

\(= \left[ {{{{x^2}} \over 2} + 6x + 11\ln |x| - {6 \over x}} \right]\left| {_1^2} \right. \) 
\( = (2 + 12 + 11\ln 2 - 3) - ({1 \over 2} + 6 - 6) \)

\(= {{21} \over 2} + 11\ln 2 \)

 d) Ta có: 

\(\begin{array}{l}
\int\limits_0^2 {\frac{1}{{{x^2} - 2x - 3}}dx = \int\limits_0^2 {\frac{1}{{\left( {x + 1} \right)\left( {x - 3} \right)}}dx} } \\
= \frac{1}{4}\int\limits_0^2 {\left( {\frac{1}{{x - 3}} - \frac{1}{{x + 1}}} \right)dx} \\
= \left. {\frac{1}{4}\left[ {\ln \left| {x - 3} \right| - \ln \left| {x + 1} \right|} \right]} \right|_0^2\\
= \frac{1}{4}\left[ { - \ln 3 - \ln 3} \right] = - \frac{1}{2}\ln 3.
\end{array}\)

 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: 

\(\eqalign{
& I = {{{\pi ^3}} \over 3} + 2\pi + {1 \over 2}\left[ {x - {{\sin 2x} \over 2}} \right]\left| {_0^{{\pi }}} \right. \cr
& = {{{\pi ^3}} \over 3} + 2\pi + {\pi \over 2} = {{2{\pi ^3} + 15\pi } \over 6}. \cr} \)