Сборник задач по высшей математике. Часть IV. Интегралы. Дифференциальные уравнения. Самохин А.В - 14 стр.

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МГТУ ГА | Москва

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14 çÌÁ×Á I. éÎÔÅÇÒÁÌØÎÏÅ ÉÓÞÉÓÌÅÎÉÅ ÆÕÎËÃÉÉ ÏÄÎÏÊ ÐÅÒÅÍÅÎÎÏÊ
ðÒÉÍÅÒ 17.
R
x
2
· cos 2x dx. ðÕÓÔØ u(x) = x
2
, v
0
(x) = cos 2x, ÔÏÇÄÁ
v
0
(x) dx = dv(x) = cos 2x dx =
1
2
cos 2x d2x =
1
2
d sin 2x.
Z
x
2
cos 2x dx =
1
2
Z
x
2
d sin 2x =
1
2
x
2
sin 2x −
Z
sin 2x dx
2
. (2)
ôÁË ËÁË dx
2
= 2x dx, ÔÏ × ÐÏÓÌÅÄÎÅÍ ÉÎÔÅÇÒÁÌÅ ÐÏÌÕÞÉÍ
Z
sin 2x dx
2
=
Z
(sin 2x) ·2x dx.
ðÏÌÁÇÁÑ u(x) = 2x, v
0
(x) = sin 2x; v
0
(x) dx = sin 2x dx =
1
2
sin 2x d2x =
= −
1
2
d cos 2x, ÉÍÅÅÍ
Z
(sin 2x) ·2x dx =
Z
2x ·
−
1
2
d cos 2x = −
Z
x d cos 2x =
= −
x cos 2x −
Z
cos 2x dx
= −x cos 2x +
1
2
Z
cos 2x d2x =
= −x cos 2x +
1
2
sin 2x + C.
ïËÏÎÞÁÔÅÌØÎÏ, ÐÏÄÓÔÁ×ÌÑÑ ÐÏÓÌÅÄÎÉÊ ÒÅÚÕÌØÔÁÔ × (2), ÐÏÌÕÞÉÍ
Z
x
2
cos 2x dx =
1
2
x
2
· sin 2x + x cos 2x −
1
2
sin 2x
+ C =
=
1
2
x
2
sin 2x +
1
2
x cos 2x −
1
4
sin 2x + C.
II. ðÒÉ ÉÎÔÅÇÒÉÒÏ×ÁÎÉÉ ÆÕÎËÃÉÊ ×ÉÄÁ: P
m
(x) arcsin ax, P
m
(x) arccos ax,
P
m
(x) arctg ax, P
m
(x) ln ax × ËÁÞÅÓÔ×Å v
0
(x) ×ÙÂÉÒÁÅÔÓÑ ÍÎÏÇÏÞÌÅÎ P
m
(x), Á
× ËÁÞÅÓÔ×Å u(x) ÏÓÔÁ×ÛÁÑÓÑ ÆÕÎËÃÉÑ.
ðÒÉÍÅÒ 18.
R
x ln x dx. ðÕÓÔØ u(x) = ln x, v
0
(x) = x, ÔÏÇÄÁ
v
0
(x) dx = x dx =
1
2
dx
2
.
Z
x ln x dx =
1
2
Z
ln x dx
2
=
1
2
x
2
ln x −
Z
x
2
d ln x
=
=
1
2
x
2
ln x −
Z
x
2
·
1
x
dx
=
1
2
x
2
ln x −
x
2
2
+ C =
=
x
2
ln x
2
−
x
2
4
+ C.
14         çÌÁ×Á I. éÎÔÅÇÒÁÌØÎÏÅ ÉÓÞÉÓÌÅÎÉÅ ÆÕÎËÃÉÉ ÏÄÎÏÊ ÐÅÒÅÍÅÎÎÏÊ

     ðÒÉÍÅÒ 17. x2 · cos 2x dx. ðÕÓÔØ u(x) = x2, v 0 (x) = cos 2x, ÔÏÇÄÁ
               R

                                             1              1
           v 0 (x) dx = dv(x) = cos 2x dx = cos 2x d2x = d sin 2x.
                                             2            Z2         
                          1                 1
        Z                   Z
          x2 cos 2x dx =      x2 d sin 2x =     x2 sin 2x − sin 2x dx2 .    (2)
                          2                 2
ôÁË ËÁË dx2 = 2x dx, ÔÏ × ÐÏÓÌÅÄÎÅÍ ÉÎÔÅÇÒÁÌÅ ÐÏÌÕÞÉÍ
                      Z             Z
                                 2
                        sin 2x dx = (sin 2x) · 2x dx.

ðÏÌÁÇÁÑ u(x) = 2x, v 0 (x) = sin 2x; v 0 (x) dx = sin 2x dx = 12 sin 2x d2x =
= − 21 d cos 2x, ÉÍÅÅÍ
                                 
                                    1
  Z                    Z                            Z
    (sin 2x) · 2x dx = 2x · −           d cos 2x = − x d cos 2x =
                                    2
                                        
                                                        1
                            Z                             Z
          = − x cos 2x − cos 2x dx = −x cos 2x +             cos 2x d2x =
                                                        2
                                                                   1
                                                    = −x cos 2x + sin 2x + C.
                                                                   2
ïËÏÎÞÁÔÅÌØÎÏ, ÐÏÄÓÔÁ×ÌÑÑ ÐÏÓÌÅÄÎÉÊ ÒÅÚÕÌØÔÁÔ × (2), ÐÏÌÕÞÉÍ
                                                       
                     1                            1
  Z
     x2 cos 2x dx =      x2 · sin 2x + x cos 2x − sin 2x + C =
                     2                            2
                                           1          1            1
                                        = x2 sin 2x + x cos 2x − sin 2x + C.
                                           2          2            4
   II. ðÒÉ ÉÎÔÅÇÒÉÒÏ×ÁÎÉÉ ÆÕÎËÃÉÊ ×ÉÄÁ: Pm (x) arcsin ax, Pm (x) arccos ax,
Pm (x) arctg ax, Pm (x) ln ax × ËÁÞÅÓÔ×Å v 0 (x) ×ÙÂÉÒÁÅÔÓÑ ÍÎÏÇÏÞÌÅÎ Pm (x), Á
× ËÁÞÅÓÔ×Å u(x) ÏÓÔÁ×ÛÁÑÓÑ       ÆÕÎËÃÉÑ.
   ðÒÉÍÅÒ 18. x ln x dx. ðÕÓÔØ u(x) = ln x, v 0 (x) = x, ÔÏÇÄÁ
                   R

                       1
 v 0 (x) dx = x dx = dx2.
                       2                                          
                            1              1
             Z                Z                         Z
                                       2        2           2
                x ln x dx =     ln x dx =      x ln x − x d ln x =
                            2              2
                                                           x2
                                                           
                1                 2 1         1
                              Z
                       2                           2
             =       x ln x − x · dx =            x ln x −      +C =
                2                    x        2            2
                                                               x2 ln x x2
                                                            =         −   + C.
                                                                  2     4

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