Variante BAC M1

Propunem spre rezolvare un exercitiu de analaiza matematica in care calculam primitiva unei functii, limita unei primitive, dar si o integrala mai complicata, astfel:

Fie functia $f:R\rightarrow R f\left(x\right)=\frac{\sin x}{1+cos^{2}x}$

a) Calculati $\int f\left(x\right) dx$

b) Fie $F:R\rightarrow R$ , o primitiva a functiei f, calcuati $\lim_{x\to 0}{\frac{F(x)-F(0)}{x^{2}}}$

c) Calcuati $\int_{0}^{2\pi} x\cdot f(x)dx$

Solutie:

a) Variante BAC M1 ! Integrala devine $\int f\left(x\right) dx=\int\frac{\sin x}{1+cos^{2}x}dx$

Ca sa rezolvam integrala folosim Metoda schimbarii de variabile. Cei care nu va mai reamintiti click aici. Astfel notam $\cos x=t$

Iar pentru a afla dx, derivam egalitatea de mai sus in functie de dx dar si in functie de dt $\left(\cos x\right)^{'} dx=t^{'} dt\Rightarrow -\sin x dx=dt\Rightarrow \sin x dx=-dt$

Astfel integrala devine $\int \frac{-dt}{1+t^{2}}=-\frac{1}{1}\arctan\frac{t}{1}=-\arctan\frac{\cos x}{1}+C=-\arctan(\cos x)+C$

Mai sus am folosit formula de la integralele uzuale $\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\frac{x}{a}+C$

b) Variante BAC M1 ! Stiind ca F este o primitiva a functie f , observam ca cu informatiile de la punctul a) stim ca $F(x)=-\arctan(\cos x)+C$ si limita devine:

$\lim\limits_{x\to 0}{\frac{F{x}-F(0)}{x^{2}}}=$

Dar mai intai calculam $F(0)=-\arctan(cos 0)=-\arctan 1=0$

Astfel limita devine $\lim\limits_{x\to 0}{\frac{-\arctan{\cos x}-0}{x^{2}}}=\lim\limits_{x\to 0}{\frac{-\arctan{\cos x}}{x^{2}}}=\frac{0}{0}$

Observati ca suntem in cazul de nedeterminare 0/0, astfel cu regula lui L’ Hospital avem ca $\lim\limits_{x\to 0}{\frac{\left(-arctan(\cos x)\right)^{'}}{\left(x^{2}\right)^{'}}}=\lim\limits_{x\to 0}{\frac{f(x)}{2x}}=\lim\limits_{x\to 0}{\frac{f^{'}(x)}{2}}=$

Mai mai intai calculam $f^{'}(x)^=\frac{\cos x\left(1+\cos^{2}x\right)-\sin x\left(-2\cos x\cdot\sin x\right)}{\left(1+\cos^{2} x\right)^{2}}=\frac{cos x+cos^{3} x+2\cos x\sin^{2} x}{\left(1+\cos^{2}x\right)}$

Pentru x=0 derivata devine $f^{'}(0)=\frac{1+1+0}{\left(1+1\right)^{2}}=\frac{2}{4}=\frac{1}{2}$

Adica limita devine: $\lim\limits_{x\to 0}{\frac{\frac{1}{2}}{2}}=\frac{1}{4}$

c) Variante BAC M1 ! Integrala devine $\int^{2\pi}_{0}x\cdot f(x)dx=\int^{2\pi}_{0}x\cdot\frac{\sin x}{1+\cos^{2}x} dx=\int^{2\pi}_{0}=\frac{x\sin x}{1+\cos^{2}x}dx=$

Pentru a rezolva integrala facem schimbarea de variabila

$t=2\pi-x\Rightarrow -x=t-2\pi\Rightarrow x=2\pi-t$

Si obtinem $(t)^{'}dt=(2\pi-x)^{'}dx\Rightarrow dt=-dx$

Iar capetele intervalului devin $x=0\Rightarrow t=2\pi-0=2\pi$

Iar pentru $x=2\pi\Rightarrow t=2\pi-2\pi=0$

Astfel integrala devine $\int^{0}_{2\pi}\left(2\pi-t\right)f\left(2\pi-t\right)\left(-dt\right)=\int_{0}^{2\pi}\left(2\pi-t\right)f\left(2\pi-t\right)dt=2\pi\int^{2\pi}_{0}f\left(2\pi-t\right)dt-\int^{2\pi}_{0}t\cdot f\left(2\pi-t\right) dt$

Dar stim ca $f\left(2\pi-t\right)=\frac{\sin(2\pi-t)}{1+\cos^{2}(2\pi-t)}$

Dar stim ca $\sin(2\pi -t)=\sin 2\pi\cdot\cos t-\cos 2\pi\sin t=-(-1)\cdot \sin t=-\sin t$ dar si $\cos(2\pi -t)=\cos 2\pi \cos t+\sin 2\pi\sin t=\cos t$ astfel $f(2\pi-t)=\frac{-\sin t}{1+\cos^{2}t}$

Si integrala devine $2\pi\int^{2\pi}_{0}\frac{-\sin t}{1+\cos^{2}t}(-dt)-\int^{2\pi}_{0}\frac{t\cdot (-\sin t)}{1+\cos^{2}t} (-dt)$

Astfel integrala devine: $\int_{0}^{2\pi}x\cdot f(x)dx=2\pi\int^{2\pi}_{0}\frac{\sin t}{1+\cos^{2}t}dt-\int^{2\pi}_{0}\frac{t\cdot \sin t}{1+\cos^{2}t} dt$
$\Rightarrow \int^{2\pi}_{0}x\cdot f(x)dx=\frac{1}{2}\cdot 2\pi\int^{2\pi}_{0}\frac{\sin t}{1+\cos^{2} t}dt=-\pi\arctan(cos t)|^{2\pi}_{0}=$

$-\pi\left(arctan(cos 2\pi)-arctan(cos 0)\right)=-\pi\left(arctan 1-arctan 1\right)=0$

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