Page 89 - MATINF Nr. 6
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            PROBLEME DE MATEMATICA PENTRU CONCURSURI                                                       89


            M 119. Fie a, b ∈ R. Calculat ,i


                      Z                      2    2
                                           (a − b ) sin 2x + 2ab cos 2x
                                                                                         dx, x ∈ I,
                         (b − a ) cos 2x + 2ab sin 2x − 4(a sin x + b cos x) + a + b + 2
                           2
                                                                              2
                                2
                                                                                   2
            I fiind un interval arbitrar pe care funct ,ia pentru care se calculeaz˘a integrala este definit˘a.
                                                                                        Daniel Jinga, Pites , ti

                                                                                           2
                                                                 2
                                                                                      2
                                                                               2
                                                                           2
            Solut ,ie.  Fie u(x) = a sin x + b cos x. Avem u (x) = a sin x + b cos x + ab sin 2x =
                                          2
              2
             b − a 2                     a + b 2
                                                                                     2
                                                         2
                                                                                2
                    cos 2x + ab sin 2x +        , deci 2u (x) − 4u(x) + 2 = (b − a ) cos 2x + 2ab sin 2x −
                2                           2
                                        2
                                   2
            4(a sin x + b cos x) + a + b + 2,   (1).
                                                                                0
                                                                                         2
                                          0
                                                                                              2
                Pe de alt˘a parte, avem u (x) = a cos x − b sin x, de unde u(x)u (x) = (a − b ) sin x cos x −
                                   2
                                  a − b 2
                                                                               2
                                                                                   2
                  2
                            2
                                                                       0
            ab sin x+ab cos x =           sin 2x+ab cos 2x, deci 2u(x)u (x) = (a −b ) sin 2x+2ab cos 2x,  (2).
                                     2
                                                          Z              0              Z         0
                                                                  2u(x)u (x)                u(x)u (x)
            Din (1) s , i (2), integrala din enunt , devine                      dx =                  dx =
                                                                2
                                                             2u (x) − 4u(x) + 2            (u(x) − 1) 2
               u(x) − 1 + 1                     1                         1
            Z                            Z                        Z
                                                        0
                               0
                                                                                    0
                             · u (x) dx =             · u (x) dx +               · u (x) dx = ln |u(x) − 1| −
                (u(x) − 1) 2                 u(x) − 1                (u(x) − 1) 2
                1                                               1
                      + C = ln |a sin x + b cos x − 1| −                   + C.
             u(x) − 1                                  a sin x + b cos x − 1
                                                    1
                                                  Z
                                                               n
            M 120. Fie a ∈ R. Calculat ,i    lim     sin(a + x ) dx.
                                             n→∞
                                                   0
               Alexandru Daniel Pˆırvuceanu s , i Cezar Alexandru Tr˘anc˘an˘au, elevi, Drobeta Turnu Severin
            Solut ,ia 1 (Quan Minh Nguyen, student, Vietnam). Avem
                                  1                             1
                               Z                           Z
                                            n                            n

                                  sin(a + x ) dx − sin a =      (sin(a + x ) − sin a) dx

                                 0                             0
                                             1
                                         Z

                                                          n
                                                                       n
                                       =     (sin a (cos x − 1) + sin x cos a dx)

                                            0
                                         1                            1
                                      Z                         Z
                                                    n                    n
                                    ≤     sin a (cos x − 1) dx +     sin x cos a dx


                                        0                            0
                                              1                              1
                                           Z                            Z
                                                                                   n
                                                     n

                                 = |sin a| ·    (cos x − 1) dx + |cos a| ·    sin x dx


                                             0                              0
                                              1                            1
                                            Z                            Z
                                                     n
                                                                                  n
                                 ≤ |sin a| ·   |cos x − 1| dx + |cos a| ·    |sin x | dx.
                                             0                            0
                                                    n 2
                                                  (x )
                       n
                              n
                                             n
            Cum |sin x | ≤ |x | s , i |1 − cos x | ≤    pentru orice x ∈ R, rezult˘a c˘a
                                                    2
                    1                     1  x 2n         1          1                 1           1
                  Z                     Z                          Z                Z
                           n
                                                                                         n
                                                                            n
                      |cos x − 1| dx ≤          dx =            s , i  |sin x | dx ≤    x dx =        .
                   0                     0   2        2(2n + 1)     0                 0          n + 1
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