MykolasD PRO +2545
Apskaičiuokite [tex]\large \int_{0}^{π}\frac{dx}{1+2^{cosx}}[/tex] [tex]\large ,[/tex] kai [tex]\large \int_{0}^{π}[/tex][tex]\large \frac{dx}{1+2^{\cos x}}=[/tex][tex]\large \int_{0}^{π}[/tex][tex]\large \frac{dx}{1+2^{\cos\left (π -x \right ) }}[/tex]
MykolasD PRO +2545
Gavoju ,kad uždavinys atitinka MVBE reikalavimus
Tomas PRO +4543
Apskaičiuokite [tex]\large\int_{0}^{\pi}\frac{dx}{1+2^{\cos x}},[/tex] kai [tex]\large\int_{0}^{\pi}\frac{dx}{1+2^{\cos x}}=\large\int_{0}^{\pi}\frac{dx}{1+2^{\cos (\pi-x)}}[/tex]
Pažymime: [tex]A=\large\int_{0}^{\pi}\frac{dx}{1+2^{\cos x}}[/tex]
$$2A=\int_{0}^{\pi}\frac{dx}{1+2^{\cos x}}+\int_{0}^{\pi}\frac{dx}{1+2^{\cos (\pi-x)}}=\int_{0}^{\pi}\frac{dx}{1+2^{\cos x}}+\int_{0}^{\pi}\frac{dx}{1+2^{-\cos x}}=\\=\int_{0}^{\pi}\left(\frac{1}{1+2^{\cos x}}+\frac{1}{1+2^{-\cos x}}\right )dx=\int_{0}^{\pi}\left(\frac{1}{1+2^{\cos x}}+\frac{2^{\cos x}}{1+2^{\cos x}}\right )dx=\\=\int_{0}^{\pi}\frac{1+2^{\cos x}}{1+2^{\cos x}}dx=\int_{0}^{\pi}dx=\pi\implies A=\dfrac{\pi}{2}$$