Question

Salut, am nevoie de ajutor la ex. 992 (explicatii pas cu pas si rezolvare completa, va rog) .
P.S din cate am inteles ar fi un ex. foarte greu .

Answer 1

Fie f o funcție de perioadă T care satisface relatia:

$f(T-x) = f(x)$

$\displaystyle \int_{0}^Tf(x) \, dx = \int_{0}^{\frac{T}{2}}f(x) \, dx + \int_{\frac{T}{2}}^Tf(t) \, dt\\ \\ \text{Inlocuim }t = T-x\text{ in a doua integrala si avem:} \\t = \dfrac{T}{2} \Rightarrow x = \dfrac{T}{2},\quad t = T \Rightarrow x = 0\\ dt =-dx \\\\ \int_{\frac{T}{2}}^0-f(T-x)\, dx = \int_{0}^{\frac{T}{2}}f(x)\, dx\\ \\ \Rightarrow \boxed{\int_{0}^Tf(x) \, dx = 2\int_{0}^{\frac{T}{2}}f(x)\, dx}$

Dar doar când T satisface relația $f(T-x) = f(x)$

$\displaystyle I =\int_0^{2\pi} \dfrac{1}{\sin^4 x+\cos^4 x}\, dx = 2\int_{0}^{\pi}\dfrac{1}{\sin^4 x+\cos^4 x}\, dx = \\\\ \text{Fiindca }2\pi \text{ e perioada pentru f si satisface acea relatie}\\ \\ = 4\int_{0}^{\frac{\pi}{2}}\dfrac{1}{\sin^4 x+\cos^4 x}\, dx = \\ \\ \text{Fiindca }\pi \text{ e perioada pentru f si satisface acea relatie} \\ \\ = 8 \int_{0}^{\frac{\pi}{4}}\dfrac{1}{\sin^4 x+\cos^4 x}\, dx\\ \\ \text{Fiindca }\dfrac{\pi}{2} \text{ e perioada pentru f si satisface acea relatie}$

Dar aici ne oprim, fiindca $\dfrac{\pi}{4}$ nu mai e perioada pentru f si nu satisface acea conditie.

$\displaystyle J = \int \dfrac{1}{\sin^4 x+\cos^4 x}\, dx = \\ \\ =\int \dfrac{1}{(\cos^2 x-\sin^2 x)^2+2\sin^2 x\cos^2 x}\, dx = \\ \\ = \int\dfrac{2}{2\cos^2 2x+\sin^2 2x}\, dx = \int \dfrac{\dfrac{2}{\cos^2 2x}}{2+\tan^2 2x}\, dx = \\ \\ = \int \dfrac{(\tan2x)'}{(\sqrt 2)^2+\tan^2 2x}\, dx = \dfrac{1}{\sqrt 2}\arctan \dfrac{\tan 2x}{\sqrt 2}+C$

$I = 8J\Big|_{0}^{\frac{\pi}{4}}=\\ \\ =\dfrac{8}{\sqrt 2}\Big(\arctan\dfrac{\tan(\frac{\pi}{4})}{\sqrt 2} - \arctan\dfrac{\tan 0}{\sqrt 2}\Big) = \\ \\ = \dfrac{8}{\sqrt 2}\Big(\arctan(\infty) - 0\Big) = \dfrac{8}{\sqrt 2}\cdot \dfrac{\pi}{2} =\boxed{2\pi \sqrt 2}$