Câu 6 trang 127 SGK Giải tích 12: Ôn tập Chương III – Nguyên hàm – Tích phân và ứng dụng...

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} )