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Calculati : \( I_1=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1+\sin x}{1+\cos x}e^{x} dx \)
\( I_2=\int_{0}^{\frac{\pi}{3}}\ln(1+sqrt3 \tan x) dx \)

P.S. Nu le-am postat la probleme pt liceu sau pt bac pt ca in respectivele sectiuni sunt 2-3 probleme si nu s-ar uita lumea daca as posta acolo si deoarece cred ca sunt peste acel nivel, desi nu s-ar incadra la mai mult de o locala.

\( I_1=\int_0^1 \frac{e^x}{1+\cos x}dx+\int_0^1\frac{sin x}{1+\cos x}e^xdx=\int_0^1 \frac{e^x}{1+\cos x}dx+\frac{sin x}{1+\cos x}e^x|_0^1-\int_0^1\frac{\cos x(1+\cos x)+\sin^2x}{(1+\cos x)^2}e^xdx= \)
\( =\int_0^1 \frac{e^x}{1+\cos x}dx+\frac{e\sin1}{1+\cos 1}-\int_0^1 \frac{e^x}{1+\cos x}dx=\frac{e\sin1}{1+\cos 1} \)

Pentru a doua integrala folosim scimbarea de variabila ,,Conservare de interval''

\( (\int_a^bf(x)dx=\int_a^bf(a+b-x)dx) \) avem

\( I_2=\int_0^{\frac{\pi}{3}}\ln\left(1+\sqrt{3}\frac{\sqrt{3}-\tan x}{1+sqrt{3}\tan x}\right)dx=\int_0^{\frac{\pi}{3}}\ln4dx-I_2 \) si atunci

\( I_2=\frac{\pi\ln 4}{6} \)