Problemă rezolvată dintre subiectele date la clasa a XII-a la ONGM 2020-2021, București.
Enunț
Fie a_n = \int _0^{\frac{\pi}{2}}\dfrac{sin^n x}{sin^nx +cos^n x} dx, \ n \in \N și l = \lim_{n\to \infty}a_n \ atunci:\\[1em] A. \ l = 0 \quad B. \ l = \dfrac{\pi}{2} \quad C.\ l = \dfrac{\pi}{4} \\[2em] D.\ a = \dfrac{\pi}{3} \quad \ E. \ l = \pi
Rezolvare
Știm că:
ctg(x)=\dfrac{cos(x)}{sin(x)}\\[1em]Atunci an devine :
a_n =\int_0^{\frac{\pi}{2}}\dfrac{sin^n x}{sin^nx +cos^n x} dx = \\[2em] \\= \int_0^{\frac{\pi}{2}}{\dfrac{1 }{\dfrac{sin^nx +cos^n x}{sin^n x}} dx} = \\[2em] \\=
\int_0^{\frac{\pi}{2}}{\dfrac{1 }{\dfrac{sin^nx}{sin^n x} + \dfrac{cos^n x}{sin^n x} } dx} = \\[2em] \\= \int_0^{\frac{\pi}{2}}{\dfrac{1 }{1 + ctg^n x}} dxȘtim că:
ctg(x)=tg(\dfrac{\pi}{2}-x)Atunci an devine :
a_n = \int_0^{\frac{\pi}{2}} {\dfrac{1 }{1 + tg^n (\dfrac{\pi}{2}-x)}}dxȘtim că:
\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dxAtunci an devine :
a_n =\int_0^{\frac{\pi}{2}} {\dfrac{1 }{1 + tg^n x}}dx \implies \\[2em]\implies a_n= \int_0^{\frac{\pi}{2}} {\dfrac{1 }{1 + \dfrac{1 }{ctg^n x}}}dx\implies \\[2em]\implies a_n= \int_0^{\frac{\pi}{2}} {\dfrac{ctg^n x }{1 + ctg^n x}}dx Dar după cum am arătat mai devreme an este:
a_n = \int_0^{\frac{\pi}{2}}{\dfrac{1 }{1 + ctg^n x}}dxAdunând ultimele două relații obținem:
2 \cdot a_n = \int_0^{\frac{\pi}{2}}{\dfrac{1 }{1 + ctg^n x}}dx + \int_0^{\frac{\pi}{2}}{\dfrac{ctg^n x }{1 + ctg^n x}}dx \implies \\[2em] \implies
2 \cdot a_n = \displaystyle \int_0^{\frac{\pi}{2}} \bigg( \dfrac{1 }{1 + ctg^n x} + \dfrac{ctg^n x }{1 + ctg^n x} \bigg)dx \implies \\[2em] \implies
2 \cdot a_n = \int_0^{\frac{\pi}{2}} \dfrac{\cancel{1 + ctg^n x}} {\cancel{1 + ctg^n x}} dx \implies \\[2em] \implies
2 \cdot a_n = \int_0^{\frac{\pi}{2}}dx \ \ \implies \\[2em] \implies 2 \cdot a_n = x\bigg |_0^\frac{\pi}{2} \implies 2\cdot a_n = \dfrac{\pi}{2} \implies \\[2em] \implies a_n = \dfrac{\pi}{4} \implies \lim_{n\to \infty}a_n = \dfrac{\pi}{4}Răspunsul corect este \textbf{C}, \dfrac{\pi}{4}
