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»CALCUL INTEGRAL

Data publicarii: 15 Iunie, 2011

ANALIZA-30

Suport teoretic:

Integrala definita, identitati trigonometrice.

Enunt:

Sa se calculeze integrala definita:

$I=\int_0^{\pi}{sinx}\cdot{sin4x}\cdot{cos3x}\cdot{dx}.$ I=\int_0^{\pi}{sinx}\cdot{sin4x}\cdot{cos3x}\cdot{dx}.

Raspuns:

I = π/4.

Rezolvare:

Folosind identitati trigonometrice cunoscute, obtinem succesiv:

$I=\int_0^{\pi}{{\frac{sin4x-sin2x}{2}}\cdot{sin4x}{dx}}=\cdots=$ I=\int_0^{\pi}{{\frac{sin4x-sin2x}{2}}\cdot{sin4x}{dx}}=\cdots= ${\frac{1}{4}}\cdot{\int_0^{\pi}{(1-cos8x){dx}}-\int_0^{\pi}{(cos2x-cos6x)}{dx}}=\cdots=$ {\frac{1}{4}}\cdot{\int_0^{\pi}{(1-cos8x){dx}}-\int_0^{\pi}{(cos2x-cos6x)}{dx}}=\cdots= $\frac{\pi}{4}.$ \frac{\pi}{4}.