Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 04 Noiembrie, 2014

EXERCITIUL 6

Suport teoretic:

Integrale trigonometrice definite,identitati trigonometrice,schimbari de variabila.

Enunt:

Sa se calculeze în 2 moduri:

$\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}.$ \int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}.

Raspuns:

$\frac{\pi\sqrt{2}}{4}.$ \frac{\pi\sqrt{2}}{4}.

Rezolvare:

Metoda I:

Fie schimbarea de variabilă

$x-\frac{\pi}{4}=t\in{[-\frac{\pi}{4},\frac{\pi}{4}]};$ x-\frac{\pi}{4}=t\in{[-\frac{\pi}{4},\frac{\pi}{4}]};

deducem de aici:

$\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}=$ \int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}= $\int_{-{\frac{\pi}{4}}}^{\frac{\pi}{4}}\frac{\sin{(t+\frac{\pi}{4})}}{cost}{dt}={\frac{\sqrt{2}}{2}}\cdot\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{{sint}+{cost}}{cost}{dt}=\cdots=\frac{\pi\sqrt{2}}{4}.$ \int_{-{\frac{\pi}{4}}}^{\frac{\pi}{4}}\frac{\sin{(t+\frac{\pi}{4})}}{cost}{dt}={\frac{\sqrt{2}}{2}}\cdot\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{{sint}+{cost}}{cost}{dt}=\cdots=\frac{\pi\sqrt{2}}{4}.

Metoda II:

Notăm

$I=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx},\;J=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}.$ I=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx},\;J=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}.

Deducem de aici, cu uşurinţă, următorul sistem:

$\begin{cases}I+J=\frac{\pi\sqrt{2}}{2}\\J-I=0\end{cases},$ \begin{cases}I+J=\frac{\pi\sqrt{2}}{2}\\J-I=0\end{cases},

din care se obţine imediat:

$I=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}=\frac{\pi\sqrt{2}}{4}.$ I=\int_{0}^{\frac{\pi}{2}}{\frac{sinx}{\cos{(x-\frac{\pi}{4})}}}{dx}=\frac{\pi\sqrt{2}}{4}.