Câu 6 trang 127 SGK Giải tích 12
Câu 6 trang 127 SGK Giải tích 12 Tính: ...
Câu 6 trang 127 SGK Giải tích 12
Tính:
Bài 6. 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{ m s} olimits} { m{inx}} + {mathop{ m cosx} olimits} )}^2}dx} )
g) (int_0^pi {{{(x + {mathop{ m s} olimits} { m{inx}})}^2}} dx)
Trả lời:
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}}
ight]} dx)
( = {1 over 4}int_0^{{pi over 2}} {(2cos 2x - cos 4x - 1)dx} )
( = {1 over 4}left[ {sin 2x - {{sin 4x} over 4} - x}
ight]_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}
ight. + ({{{2^x}} over {ln 2}} + {{{2^{ - x}}} over {ln 2}})left| {_0^1}
ight. )(= {1 over {ln 2}} )
c)
(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 + 11ln |x| - {6 over x}}
ight]left| {_1^2}
ight. )
( = (2 + 12 + 11ln 2 - 3) - ({1 over 2} + 6 - 6) )
(= {{21} over 2} + 11ln 2 )
d)
(eqalign{
& int_0^2 {{1 over {{x^2} - 2x - 3}}} dx = int_0^2 {{1 over {(x + 1)(x - 3)}}dx = {1 over 4}} int_0^2 {({1 over {x - 3}} - {1 over {x + 1}})dx} cr
& = {1 over 4}left[ {ln |x - 3| - ln |x + 1|}
ight]left| {_0^2}
ight. = {1 over 4}left[ {- ln 3 - ln 3}
ight] cr
& = {-1 over 2} ln 3cr} )
e)
(eqalign{
& int_0^{{pi over 2}} {{{({mathop{
m s}
olimits} {
m{inx}} + {mathop{
m cosx}
olimits} )}^2}dx} = int_0^{{pi over 2}} {(1 + sin 2x)dx} cr
& = left[ {x - {{cos 2x} over 2}}
ight]left| {_0^{{pi over 2}}}
ight. = {pi over 2} + 1 cr} )
g)
(eqalign{
& I = int_0^pi {{{(x + {mathop{
m s}
olimits} {
m{inx)}}}^2}} dxint_0^pi {({x^2}} + 2xsin x + {sin ^2}x)dx cr
& = left[ {{{{x^3}} over 3}}
ight]left| {_0^pi }
ight. + 2int_0^pi {xsin xdx + {1 over 2}} int_0^pi {(1 - cos 2x)dx} cr} )
Tính :(J = int_0^pi {xsin xdx} )
Đặt (u = x ⇒ u’ = 1) và (v’ = sinx ⇒ v = -cos x)
Suy ra:
(J = left[ { - x{mathop{ m cosx} olimits} } ight]left| {_0^pi } ight. + int_0^pi {{mathop{ m cosxdx} olimits} = pi + left[ {{mathop{ m s} olimits} { m{inx}}} ight]} left| {_0^pi } ight. = pi )
Do đó:
(eqalign{
& I = {{{pi ^3}} over 3} + 2pi + {1 over 2}left[ {x - {{sin 2x} over 2}}
ight]left| {_0^{{pi }}}
ight. cr
& = {{{pi ^3}} over 3} + 2pi + {pi over 2} = {{2{pi ^3} + 15pi } over 6} cr} )
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