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